Class 12 Maths Solutions Vector Algebra
In our daily life, we generally come across two types of quantities, namely scalars and vectors.
Scalars A quantity that has magnitude only is known as a scalar.
Examples Each of the quantities mass, length, time, temperature, density, speed, etc., is a scalar.
Vectors A quantity that has magnitude as well as direction is called a vector.
Examples Each of the quantities force, velocity, acceleration and momentum is a vector.
However, we define a vector as given below.
‘A directed line segment is called a vector’.
A directed line segment with initial point A and the terminal point B, is the vector denoted by \(\overrightarrow{A B}\).
The magnitude of \(\overrightarrow{A B}\) is denoted by |\(\overrightarrow{A B}\)|.
Remark We usually denote a vector by a single letter with an arrow on it and its magnitude is denoted by this letter only.
Thus, \(\overrightarrow{A B}\) = \(\vec{a}\) and |\(\overrightarrow{A B}\)| = |\(\vec{a}\)| = a.
Unit Vector A vector \(\vec{a}\) is called a unit vector if |\(\vec{a}\)| = 1 and it is denoted by \(\hat{a}\) .
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Equal Vectors Two vectors \(\vec{a}\) and \(\vec{b}\) are said to be equal if they have the same magnitude and the same direction regardless of the positions of their initial points.
Negative Of A Vector A vector having the same magnitude as that of a given vector \(\vec{a}\) and the direction opposite to that of \(\vec{a}\) is called the negative of \(\vec{a}\), to be denoted by –\(\vec{a}\).
Thus, if \(\overrightarrow{A B}\) = \(\vec{a}\), then \(\overrightarrow{B A}\) = –\(\vec{a}\).
Zero Or Null Vector A vector whose initial and terminal points coincide is called a zero vector, denoted by \(\overrightarrow{0}\).
Clearly, the magnitude of a zero vector is 0 but it cannot be assigned a definite direction.
Thus, \(\overrightarrow{A A}\) = \(\overrightarrow{0}\).
Coinitial Vectors Two or more vectors having the same initial point are called coinitial vectors.
\(\overrightarrow{O A}\) and \(\overrightarrow{O B}\) are the two coinitial vectors having the same initial point O.
Collinear Vectors Vectors having the same or parallel supports are known as collinear vectors.
In the given figure \(\overrightarrow{A B}\), \(\overrightarrow{B C}\) and \(\overrightarrow{A C}\) are collinear vectors.
Like Vectors Collinear vectors having the same direction are called like vectors.
Thus, \(\overrightarrow{A B}\), \(\overrightarrow{B C}\) and \(\overrightarrow{A C}\) shown above are like vectors.
Unlike Vectors Collinear vectors having opposite directions are known as unlike vectors.
In the given figure PQ ∥ RS.
∴ \(\overrightarrow{P Q}\) and \(\overrightarrow{S R}\) are unlike vectors.
Free Vectors If the initial point of a vector is not specified then it is said to be a free vector.
Localised Vectors A vector drawn parallel to a given vector through a specified point as the initial point is called a localised vector.
Coplanar Vectors Three or more nonzero vectors lying in the same plane or parallel to the same plane are said to be coplanar, otherwise they are called noncoplanar.
Position Vector Of A Point Let O be the origin and let A be a point such that \(\overrightarrow{O A}\) = \(\vec{a}\), then we say that the position vector of A is \(\vec{a}\).
Solved Examples
Example 1 Classify the following measures as scalars and vectors:
(1) 30 seconds
(2) 100 m2
(3) 50 km/hr
(4) 20 gm/cm3
(5) 26 m/s towards south
Solution
(1) 30 seconds represents time, which is scalar.
(2) 100 m2 represents an area, which is scalar.
(3) 50 km/hr represents speed, which is scalar.
(4) 20 gm/cm3 represents density, which is scalar.
(5) 26 m/s towards south represents velocity, which is a vector.
Example 2 Represent graphically a displacement of 50 km, 60° west of north.
Solution
The vector \(\overrightarrow{O A}\) given below represents a displacement of 50 km, 60° west of north.
Vector Addition
Let \(\vec{a}\) and \(\vec{b}\) be any two vectors. Take any point O and draw segments \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) such that \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{A B}\) = \(\vec{b}\). Join OB. Then, \(\overrightarrow{O B}\) is called the sum or resultant of \(\vec{a}\) and \(\vec{b}\).
∴ \((\vec{a}+\vec{b})=\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}\)
Triangle Law Of Addition Of Forces
In a △OAB, if \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) represent \(\vec{a}\) and \(\vec{b}\) respectively, then \(\overrightarrow{O B}\) represents \((\vec{a}+\vec{b})\).
This is known as Triangle Law of Addition of Forces.
Parallelogram Law Of Addition Of Forces In a ∥gm OABC, if \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) represent \(\vec{a}\) and \(\vec{b}\) respectively, then \(\overrightarrow{O B}\) represents \((\vec{a}+\vec{b})\).
This is known as Parallelogram Law of Addition of Forces.
Laws of Addition of Vectors
Theorem 1 (Commutative Law) Vector addition is commutative, i.e., \(\vec{a}+\vec{b}=\vec{b}+\vec{a} .\)
Proof
Let \(\vec{a}\) and \(\vec{b}\) be the given vectors represented by \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) respectively. Complete the parallelogram OABC.
Then, \(\overrightarrow{O C}=\overrightarrow{A B}=\vec{b}\)
and \(\overrightarrow{C B}=\overrightarrow{O A}=\vec{a}\)
∴ \(\vec{a}+\vec{b}=\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}\)
and \(\vec{b}+\vec{a}=\overrightarrow{O C}+\overrightarrow{C B}=\overrightarrow{O B}\).
Hence, \(\vec{a}+\vec{b}=\vec{b}+\vec{a} .\)
WBBSE Class 12 Vector Algebra Solutions
Theorem 2 (Associative Law) Vector addition is associative, i.e., \((\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})\).
Proof
Let \(\overrightarrow{O A}\) = \(\vec{a}\), \(\overrightarrow{A B}\) = \(\vec{b}\) and \(\overrightarrow{B C}\) = \(\vec{c}\).
Join OB, OC and AC.
\((\vec{a}+\vec{b})+\vec{c}=(\overrightarrow{O A}+\overrightarrow{A B})+\overrightarrow{B C}\)= \((\overrightarrow{O B}+\overrightarrow{B C})\) [∵ \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) = \(\overrightarrow{O B}\)]
= \(\overrightarrow{O C}\)
\(\vec{a}+(\vec{b}+\vec{c})=\overrightarrow{O A}+(\overrightarrow{A B}+\overrightarrow{B C})\)= \((\overrightarrow{O A}+\overrightarrow{A C})\) [∵ \(\overrightarrow{A B}\) + \(\overrightarrow{B C}\) = \(\overrightarrow{A C}\)]
= \(\overrightarrow{O C}\)
∴ \((\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})\).
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Theorem 3 (Existence of Additive Identity) For any vector \(\vec{a}\), prove that \(\vec{a}+\overrightarrow{0}=\overrightarrow{0}+\vec{a}=\vec{a} .\)
Proof
Let \(\overrightarrow{O A}\) = \(\vec{a}\). Then, \(\vec{a}+\overrightarrow{0}=\overrightarrow{O A}+\overrightarrow{A A}=\overrightarrow{O A}=\vec{a}\)
and, \(\overrightarrow{0}+\vec{a}=\overrightarrow{O O}+\overrightarrow{O A}=\overrightarrow{O A}=\vec{a} .\)
∴ \(\vec{a}+\overrightarrow{0}=\overrightarrow{0}+\vec{a}=\vec{a} .\)
Remark The vector \(\overrightarrow{0}\) is called the additive identity for vectors.
Theorem 4 (Existence of Additive Inverse) For any vector \(\vec{a}\), prove that \(\vec{a}+(-\vec{a})=(-\vec{a})+\vec{a}=\overrightarrow{0}\).
Proof
Let \(\overrightarrow{O A}\) = \(\vec{a}\). Then, \(\overrightarrow{A o}\) = –\(\vec{a}\).
∴ \(\vec{a}+(-\vec{a})=\overrightarrow{O A}+\overrightarrow{A O}=\overrightarrow{O O}=\overrightarrow{0}\)
and, \((-\vec{a})+\vec{a}=\overrightarrow{A O}+\overrightarrow{O A}=\overrightarrow{A A}=\overrightarrow{0}\)
Hence, \(\vec{a}+(-\vec{a})=(-\vec{a})+\vec{a}=\overrightarrow{0}\).
Remark The vector –\(\vec{a}\) is called the additive inverse of \(\vec{a}\).
Difference Of Two Vectors For any two vectors \(\vec{a}\) and \(\vec{b}\), we define \(\vec{a}-\vec{b}=\vec{a}+(-\vec{b})\).
Let \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{O B}\) = \(\vec{b}\). Then, \(\overrightarrow{B O}\) = –\(\vec{b}\).
∴ \((\vec{a}-\vec{b})=\vec{a}+(-\vec{b})\)
= \(\overrightarrow{O A}+\overrightarrow{B O}=\overrightarrow{B O}+\overrightarrow{O A}=\overrightarrow{B A}\)
Thus, (\(\overrightarrow{O A}\) – \(\overrightarrow{O B}\)) = \(\overrightarrow{B A}\).
Similarly, (\(\overrightarrow{O B}\) – \(\overrightarrow{O A}\)) = \(\overrightarrow{A B}\).
Scalar Multiplication of a Vector
The scalar multiple of \(\vec{a}\) by a scalar k is the vector k \(\vec{a}\) such that
(1) \(|k \vec{a}|=|k||\vec{a}|\)
(2) direction of k \(\vec{a}\) is the same as that of \(\vec{a}\), when k > 0 and opposite to that of \(\vec{a}\) when k < 0.
Examples (1) 5 \(\vec{a}\) is the vector whose magnitude is 5 times the magnitude of \(\vec{a}\) and whose direction is the same as that of \(\vec{a}\).
(2) -2 \(\vec{a}\) is the vector whose magnitude is 2 times the magnitude of \(\vec{a}\) and whose direction is opposite to that of \(\vec{a}\).
Components of a Vector
Let O be the origin and let P(x,y,z) be any point in space. Let \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) be unit vectors along the x-axis, y-axis and z-axis respectively. Let the position vector of P be \(\vec{r}\).
Then, \(\vec{r}\) = \((x \hat{i}+y \hat{j}+z \hat{k}) .\)
This form of a vector is called its component form.
Here x, y, z are called the scalar components of \(\vec{r}\) and \(x \hat{i}, y \hat{j}, z \hat{k}\) are called its vector components.
Also, \(|\vec{r}\ = |x \hat{i}+y \hat{j}+z \hat{k}| = \sqrt{x^2+y^2+z^2}\).
Direction Ratios and Direction Cosines of a vector
Consider a vector \(\vec{r} = a \hat{i}+b \hat{j}+c \hat{k}\).
Then, the numbers a, b, c are called the direction ratios of \(\vec{r}\).
Direction cosines of \(\vec{r}\) are given by
\(\frac{a}{\sqrt{a^2+b^2+c^2}} \frac{b}{\sqrt{a^2+b^2+c^2}} \text { and } \frac{c}{\sqrt{a^2+b^2+c^2}} \text {. }\)Note If l, m, n are the direction cosines of a vector then we always have (l2 + m2 + n2) = 1.
Remark If A(x1, y1, z1) and B(x2, y2, z2) be any two points in space, then direction ratios of \(\overrightarrow{A B}\) are (x2 – x1), (y2 – y1), (z2 – z1) and direction cosines of \(\overrightarrow{A B}\) are:
\(\frac{\left(x_2-x_1\right)}{r}, \frac{\left(y_2-y_1\right)}{r}, \frac{\left(z_2-z_1\right)}{r}\),
where r = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2} .\)
Solved Examples
Example 1 Let \(\vec{a} = a_1 \hat{i}+3 \hat{j}+a_3 \hat{k}\) and \(\vec{b} = 2 \hat{i}+b_2 \hat{j}+\hat{k}\). If \(\vec{a}\) = \(\vec{b}\), find the values of a1, b2 and a3.
Solution
\(\vec{a}\) = \(\vec{b}\) ⇔ \(a_1 \hat{i}+3 \hat{j}+a_3 \hat{k}=2 \hat{i}+b_2 \hat{j}+\hat{k}\)
⇔ a1 = 2, b2 = 3, a3 = 1.
The values of a1, b2 and a3 are 2,3,1.
Example 2 Let \(\vec{a}\) = \(3 \hat{i}+2 \hat{j}\) and \(\vec{b}\) = \(2 \hat{i}+3 \hat{j}\). Is |\(\vec{a}\)| = |\(\vec{b}\)|? Is \(\vec{a}\) = \(\vec{b}\)?
Solution
We have
\(|\vec{a}|=\sqrt{3^2+2^2}=\sqrt{13} \text { and }|\vec{b}|=\sqrt{2^2+3^2}=\sqrt{13} \text {. }\)∴ |\(\vec{a}\)| = |\(\vec{b}\)|.
But, \(3 \hat{i}+2 \hat{j}\) ≠ \(2 \hat{i}+3 \hat{j}\) and therefore, \(\vec{a}\) ≠ \(\vec{b}\).
Example 3 Find a unit vector in the direction of the vector \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)
Solution
\(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) ⇒ \(|\vec{a}|=\sqrt{1^2+2^2+3^2}=\sqrt{14} \text {. }\)
Unit vector in the direction of \(\vec{a}\) is given by
\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{(\hat{i}+2 \hat{j}+3 \hat{k})}{\sqrt{14}}=\left(\frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\right) .\)Example 4 Find a unit vector in the direction of \(\overrightarrow{A B}\), where A(1,2,3) and B(4, 5, 6) are the given points.
Solution
We have
p.v. of A = \((\hat{i}+2 \hat{j}+3 \hat{k})\) and p.v. of B = \((4 \hat{i}+5 \hat{j}+6 \hat{k})\)
∴ \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \((4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=(3 \hat{i}+3 \hat{j}+3 \hat{k})\), and
\(|\overrightarrow{A B}| = \sqrt{3^2+3^2+3^2}=\sqrt{27}\)∴ unit vector in the direction of \(\overrightarrow{A B}=\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}\)
= \(\frac{(3 \hat{i}+3 \hat{j}+3 \hat{k})}{\sqrt{27}}=\frac{3(\hat{i}+\hat{j}+\hat{k})}{3 \sqrt{3}}=\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}\)
= \(\left(\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\right) .\)
Example 5 Find a vector in the direction of the vector \(\vec{a}=(3 \hat{i}+\hat{j})\) that has magnitude 5 units.
Solution
\(\vec{a}=(3 \hat{i}+\hat{j})\) ⇒ \(|\vec{a}| = \sqrt{3^2+1^2}=\sqrt{10}\)
Unit vector in the direction of \vec{a} is given by
\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{(3 \hat{i}+\hat{j})}{\sqrt{10}}=\left(\frac{3}{\sqrt{10}} \hat{i}+\frac{1}{\sqrt{10}} \hat{j}\right) .\)Hence, the required vector, 5 \(\hat{a}\) = \(\left(\frac{15}{\sqrt{10}} \hat{i}+\frac{5}{\sqrt{10}} \hat{j}\right)\)
Example 6 Find a vector in the direction of the vector \(3 \hat{i}-\hat{j}+4 \hat{k}\), which has magnitude 6 units.
Solution
Let \(\vec{a}\) = \((3 \hat{i}-\hat{j}+4 \hat{k})\). Then,
|\(\vec{a}\)| = \(\sqrt{3^2+(-1)^2+4^2}=\sqrt{26}\)
∴ unit vector, \(\hat{a}=\left\{\frac{3}{\sqrt{26}} \hat{i}-\frac{1}{\sqrt{26}} \hat{j}+\frac{4}{\sqrt{26}} \hat{k}\right\}\)
Hence, the required vector, \(6 \hat{a}=\left\{\frac{18}{\sqrt{26}} \hat{i}-\frac{6}{\sqrt{26}} \hat{j}+\frac{24}{\sqrt{26}} \hat{k}\right\}\).
Step-by-Step Solutions to Vector Algebra Problems
Example 7 If \(\vec{a}=(\hat{i}+2 \hat{j}-3 \hat{k}) \text { and } \vec{b}=(2 \hat{i}+3 \hat{j}+\hat{k})\), find a unit vector in the direction of \((\vec{a}+\vec{b})\).
Solution
Let \(\vec{c}=(\vec{a}+\vec{b})\). Then,
\(\vec{c}=(\vec{a}+\vec{b})\)= \((\hat{i}+2 \hat{j}-3 \hat{k})+(2 \hat{i}+3 \hat{j}+\hat{k})\)
= \((3 \hat{i}+5 \hat{j}-2 \hat{k})\)
∴ \(|\vec{c}|=\sqrt{3^2+5^2+(-2)^2}=\sqrt{38} .\)
Hence, the required vector is given by
\(\hat{c}=\frac{\vec{c}}{|\vec{c}|}=\frac{(3 \hat{i}+5 \hat{j}-2 \hat{k})}{\sqrt{38}}=\left(\frac{3}{\sqrt{38}} \hat{i}+\frac{5}{\sqrt{38}} \hat{j}-\frac{2}{\sqrt{38}} \hat{k}\right) .\)Example 8 Write two different vectors having same magnitude.
Solution
Consider the vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-4 \hat{k} \text { and } \vec{b}=4 \hat{i}+2 \hat{j}+3 \hat{k}\)
Clearly, \(\vec{a}\) ≠ \(\vec{b}\).
But, \(|\vec{a}|=\sqrt{2^2+3^2+(-4)^2}=\sqrt{29} \text { and }|\vec{b}|=\sqrt{4^2+2^2+3^2}=\sqrt{29} \text {. }\)
Thus, |\(\vec{a}\)| = |\(\vec{b}\)| and \(\vec{a}\) ≠ \(\vec{b}\).
Example 9 Write two different vectors having same direction.
Solution
Clearly, 3 \(\overrightarrow{A B}\) and 5 \(\overrightarrow{A B}\) are two different vectors having the same direction.
Example 10 Find the scalar and vector components of the vector with initial point A(3,2) and terminal point B(-5,7).
Solution
Given vector is \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \((-5 \hat{i}+7 \hat{j})-(3 \hat{i}+2 \hat{j})=(-8 \hat{i}+5 \hat{j})\)
Vector components of \(\overrightarrow{A B}\) are \(-8 \hat{i} and 5 \hat{j}\).
Scalar components of \(\overrightarrow{A B}\) are -8 and 5.
Example 11 If A(1,2,-3) and B(-1,-2,1) are two given points in space then find (1) the direction ratios of \(\overrightarrow{A B}\) and (2) the direction cosines of \(\overrightarrow{A B}\).
Solution
(1) DRs of \(\overrightarrow{A B}\) are (-1-1), (-2-2), (1+3), i.e., -1, -4, 4.
(2) r2 = {(-2)2 + (-4)2 + 42} = 36 ⇒ r = √36 = 6.
Hence, DCs of \(\overrightarrow{A B}\) are \(\frac{-2}{6}, \frac{-4}{6}, \frac{4}{6} \text {, i.e., } \frac{-1}{3}, \frac{-2}{3}, \frac{2}{3} \text {. }\)
Example 12 Find the direction ratios and the direction cosines of the vector \(\vec{r}=2 \hat{i}+3 \hat{j}+\hat{k}\).
Solution
Given vector is \(\vec{r}=2 \hat{i}+3 \hat{j}+\hat{k}\).
DRs of \(\vec{r}\) are 2, 3, 1.
Also, \(|\vec{r}|=\sqrt{2^2+3^2+1^2}=\sqrt{14} \text {. }\)
So, DCs of \(\vec{r}\) are \(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}\)
Secion Formulae
Theorem 1 (Section Formula for Internal Division) Let A and B be two points with position vectors \(\vec{a}\) and \(\vec{b}\) respectively and let P be a point dividing AB internally in the ratio m:n. Let \(\overrightarrow{O P}\) = \(\vec{r}\). Then, prove that
\(\vec{r}=\frac{(m \vec{b}+n \vec{a})}{(m+n)}\)Proof
Let O be the origin. Then, \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{O B}\) = \(\vec{b}\).
Let P be a point on AB such that \(\frac{A P}{P B}=\frac{m}{n}\). Then,
\(\frac{A P}{P B}=\frac{m}{n}\)⇒ n.AP = m.PB
⇒ \(n(\overrightarrow{A P})=m(\overrightarrow{P B})\)
⇒ \(n(\overrightarrow{O P}-\overrightarrow{O A})=m(\overrightarrow{O B}-\overrightarrow{O P})\)
⇒ \((m+n) \overrightarrow{O P}=m \overrightarrow{O B}+n \overrightarrow{O A}=m \vec{b}+n \vec{a}\)
⇒ \(\overrightarrow{O P}=\frac{m \vec{b}+n \vec{a}}{(m+n)} ⇒ \vec{r}=\frac{(m \vec{b}+n \vec{a})}{(m+n)}\)
Corollary The position vector of the midpoint of the join of two points with position vectors \(\vec{a}\) and \(\vec{b}\) is \(\frac{1}{2}(\vec{a}+\vec{b}) .\)
Proof
Let A and B be two points with position vectors \(\vec{a}\) and \(\vec{b}\) respectively.
Let P be the midpoint of AB.
Then, P divides AB in the ratio 1:1.
∴ \(\overrightarrow{O P}=\frac{(1 \cdot \vec{b}+1 \cdot \vec{a})}{(1+1)}=\frac{1}{2}(\vec{a}+\vec{b})\)
Theorem 2 (Section Formula for External Division) Let A and B be two points with position vectors \(\vec{a}\) and \(\vec{b}\) respectively and let P be a point dividing AB externally in the ratio m:n. Let \(\overrightarrow{O P}\) = \(\vec{r}\). Then, prove that
\(\vec{r}=\frac{(m \vec{b}-n \vec{a})}{(m-n)}\)Proof
Let O be the origin. Then, \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{O B}\) = \(\vec{b}\).
Let AB be produced to P such that AP:BP = m:n.
Now, \(\frac{A P}{B P}=\frac{m}{n}\)
⇒ n.AP = m.BP ⇒ n.\(\overrightarrow{A P}\) = m.\(\overrightarrow{B P}\)
⇒ \(n(\overrightarrow{O P}-\overrightarrow{O A})=m(\overrightarrow{O P}-\overrightarrow{O B})\)
⇒ \((m-n) \overrightarrow{O P}=(m \overrightarrow{O B}-n \overrightarrow{O A})=(m \vec{b}-n \vec{a})\)
⇒ \(\overrightarrow{O P}=\frac{(m \vec{b}-n \vec{a})}{(m-n)} ⇒ \vec{r}=\frac{(m \vec{b}-n \vec{a})}{(m-n)}\)
Real-Life Applications of Vector Algebra Concepts
Example 13 Find the position vector of a point R which divides the line joining the points \(P(\hat{i}+2 \hat{j}-\hat{k}) \text { and } Q(-\hat{i}+\hat{j}+\hat{k})\) in the ratio 2:1 (1) internally and (2) externally.
Solution
Here \(\vec{a}=(\hat{i}+2 \hat{j}-\hat{k}) \text { and } \vec{b}=(-\hat{i}+\hat{j}+\hat{k})\). Also, m = 2, n = 1.
(1) When R divides PQ internally in the ratio 2:1
Then, p.v. of R = \(\frac{(m \vec{b}+n \vec{a})}{(m+n)}\)
= \(\frac{2(-\hat{i}+\hat{j}+\hat{k})+1 \cdot(\hat{i}+2 \hat{j}-\hat{k})}{(2+1)}=\frac{(-\hat{i}+4 \hat{j}+\hat{k})}{3} .\)
(2) When R divides PQ externally in the ratio 2:1
Then, p.v. of R = \(\frac{(m \vec{b}-n \vec{a})}{(m-n)}\)
= \(\frac{2(-\hat{i}+\hat{j}+\hat{k})-1 \cdot(\hat{i}+2 \hat{j}-\hat{k})}{(2-1)}=(-3 \hat{i}+3 \hat{k})\)
Example 14 Find the position vector of the midpoint of the vector joining the point \(A(2 \hat{i}+3 \hat{j}+4 \hat{k}) \text { and } B(4 \hat{i}+\hat{j}-2 \hat{k})\)
Solution
The position vectors of A and B are given by
\(\vec{a}=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \text { and } \vec{b}=(4 \hat{i}+\hat{j}-2 \hat{k})\)∴ p.v. of midpoint of AB
= \(\frac{1}{2}(\vec{a}+\vec{b})\)
= \(\frac{1}{2}\{(2 \hat{i}+3 \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})\}\)
= \((3 \hat{i}+2 \hat{j}+\hat{k})\).
The position vector of the midpoint of the vector joining the point = \((3 \hat{i}+2 \hat{j}+\hat{k})\).
Example 15 Show that the points A, B, C with position vectors \(\vec{a}=(3 \hat{i}-4 \hat{j}-4 \hat{k}), \vec{b}=(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \vec{c}=(\hat{i}-3 \hat{j}-5 \hat{k})\) respectively, from the vertices of a right-angled triangle.
Solution
We have
\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A) = \((\vec{b}-\vec{a})\)
= \((2 \hat{i}-\hat{j}+\hat{k})-(3 \hat{i}-4 \hat{j}-4 \hat{k})=(-\hat{i}+3 \hat{j}+5 \hat{k})\);
\(\overrightarrow{B C}\) = (p.v. of C)-(p.v. of B) = \((\vec{c}-\vec{b})\)
= \((\hat{i}-3 \hat{j}-5 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k})=(-\hat{i}-2 \hat{j}-6 \hat{k})\);
\(\overrightarrow{C A}\) = (p.v. of A)-(p.v. of C) = \((\hat{a}-\hat{c})\)
= \((3 \hat{i}-4 \hat{j}-4 \hat{k})-(\hat{i}-3 \hat{j}-5 \hat{k})=(2 \hat{i}-\hat{j}+\hat{k})\)
∴ \(|\overrightarrow{A B}|^2=\left\{(-1)^2+3^2+5^2\right\}=35\)
\(|\overrightarrow{A B}|^2=\left\{(-1)^2+3^2+5^2\right\}=35\)and \(|\overrightarrow{C A}|^2=\left\{2^2+(-1)^2+1^2\right\}=6\)
Thus, \(|\overrightarrow{A B}|^2+|\overrightarrow{C A}|^2=|\overrightarrow{B C}|^2\), i.e., AB2 + CA2 = BC2.
Hence, △ABC is right angled at A.
Scalar Product of Vectors
Angle Between Two Vectors Let \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\) be two given vectors. Take any point O, draw OA ∥ PQ and OB ∥ RS. Then, ∠AOB = θ is called the angle between \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\), provided 0 ≤ θ ≤ π.
If θ = 0 or θ = π then \(\overrightarrow{P Q}\) ∥ \(\overrightarrow{R S}\).
If θ = \(\frac{\pi}{2}\) then \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\) are called perpendicular or orthogonal vectors.
Scalar Product Or Dot Product Of Two Vectors Let \(\vec{a}\) and \(\vec{b}\) be two vectors and let θ be the angle between them. Then, the scalar product, or dot product, of \(\vec{a}\) and \(\vec{b}\), denoted by \(\vec{a}\).\(\vec{b}\), is defined as \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\) cos θ = ab cos θ.
Clearly, the scalar product of two vectors is a scalar.
Types of Vectors Explained
Angle Between Two Vectors In Terms Of Scalar Product Let θ be the angle between two nonzero vectors \(\vec{a}\) and \(\vec{b}\). Then,
\(\vec{a}\).\(\vec{b}\) = \(|\vec{a}||\vec{b}| \cos \theta \text {. }\)
∴ \(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \Rightarrow \theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right) .\)
Length Of A Vector Let \(\vec{a}\) be any vector.
Then, \(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \Rightarrow \theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right) .\)
∴ \(|\vec{a}|=\sqrt{\vec{a} \cdot \vec{a}} .\)
Remarks (1) If \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\), we define \(\vec{a}\).\(\vec{b}\) = 0.
(2) If \(\vec{a}\) and \(\vec{b}\) are like vectors, we have θ = 0.
∴ \(\vec{a} \cdot \vec{b}=a b \cos 0^{\circ}=a b .\)
(3) If \(\vec{a}\) and \(\vec{b}\) are unlike vectors, we have θ = π.
∴ \(\vec{a} \cdot \vec{b}=a b \cos \pi=-a b .\)
(4) If \(\vec{a}\) and \(\vec{b}\) are orthogonal vectors, we have θ = \(\frac{\pi}{2}\).
∴ \(\vec{a} \cdot \vec{b}=a b \cos \frac{\pi}{2}=0 .\)
Example 1 Let \(\vec{a}\) and \(\vec{b}\) be two given vectors such that |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and the angle between them is 60°. Find \(\vec{a}\).\(\vec{b}\).
Solution
Clearly, we have
\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos 60^{\circ}=\left(3 \times 4 \times \frac{1}{2}\right)=6 \text {. }\)Example 2 Let \(\vec{a}\) and \(\vec{b}\) be two given vectors such that |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 1 and \(\vec{a}\).\(\vec{b}\) = 1. Find the angle between \(\vec{a}\) and \(\vec{b}\).
Solution
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,
\(\vec{a} \cdot \vec{b}=1 \Rightarrow|\vec{a}||\vec{b}| \cos \theta=1\)⇒ (2 x 1)cos θ = 1 [∵ |\(\vec{a}\)| = 2 and |\(\vec{b}\)| = 1]
⇒ \(\cos \theta=\frac{1}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)
Hence, the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{3}\).
Orthonormal Vector Triad Let \(\hat{i}, \hat{j}, \hat{k}\) be unit vectors along three mutually perpendicular coordinate axes, the x-axis, y-axis and z-axis respectively. Then, these vectors are said to form an orthonormal triad of vectors.
Clearly, we have
(1) \(\hat{i} \cdot \hat{i}=|\hat{i}||\hat{i}| \cos 0^{\circ}=1\)
Similarly, \(\hat{j} \cdot \hat{j}=1 \text { and } \hat{k} \cdot \hat{k}=1 \text {. }\)
Thus, \(\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\).
(2) \(\hat{i} \cdot \hat{j}=|\hat{i}||\hat{j}| \cos \frac{\pi}{2}=0 \text {. }\)
Similarly, \(\hat{i} \cdot \hat{k}=0, \hat{j} \cdot \hat{k}=0, \hat{j} \cdot \hat{i}=0, \hat{k} \cdot \hat{i}=0 \text { and } \hat{k} \cdot \hat{j}=0\)
∴ \(\hat{i} \cdot \hat{j}=\hat{i} \cdot \hat{k}=\hat{j} \cdot \hat{i}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=\hat{k} \cdot \hat{j}=0\)
Projection Of \(\vec{b}\) on \(\vec{a}\) Let \(\overrightarrow{O A}=\vec{a}\), \(\overrightarrow{O B}=\vec{b}\) and ∠AOB = θ. Draw BM ⊥ OA. Then, OM is projection of \(\vec{b}\) on \(\vec{a}\).
OM = projection of \(\vec{b}\) on \(\vec{a}\) = \(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|}\)
Properties of Scalar Product
Theorem 1 (Commutative Law) Prove that \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)
Proof
If \(\vec{a}\) or \(\vec{b}\) is a zero vector then \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{b}\).\(\vec{a}\) = 0.
So, in this case, \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)
Now, let \(\vec{a}\) and \(\vec{b}\) be any two nonzero vectors, and let θ be the angle between them. Then,
\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=a b \cos \theta \text {, }\) and
\(\vec{b} \cdot \vec{a}=|\vec{b}||\vec{a}| \cos (-\theta)=b a \cos \theta=a b \cos \theta\)So, in this case also, \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)
Hence, \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)
Theorem 2 Prove that \(\vec{a}\).\(\vec{b}\) = 0 ⇔ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\).
Proof
Let \(\vec{a}\).\(\vec{b}\) = 0.
Then, \(\vec{a}\).\(\vec{b}\) = 0 ⇒ ab cos θ = 0
⇒ a = 0 or b = 0 or cos θ = 0
⇒ a = 0 or b = 0 or θ = \(\frac{\pi}{2}\)
⇒ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\).
Thus, \(\vec{a}\).\(\vec{b}\) = 0 ⇒ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\)
Conversely, let \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\).
If \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) then by definition, \(\vec{a}\).\(\vec{b}\) = 0
If \(\vec{a}\) ⊥ \(\vec{b}\) then \(\vec{a}\).\(\vec{b}\) = ab cos \(\frac{\pi}{2}\) = 0.
∴ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) ⇔ \(\vec{a}\) ⊥ \(\vec{b}\) ⇒ \(\vec{a}\).\(\vec{b}\) = 0.
Hence, \(\vec{a}\).\(\vec{b}\) = 0. ⇔ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\).
Theorem 3 (Cauch Schwartz Inequality) Prove that \(|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}|\).
Proof
When \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\), then
\(|\vec{a} \cdot \vec{b}|=0=|\vec{a}||\vec{b}| \text {. }\)So, let us assume that \(\vec{a}\) ≠ \(\overrightarrow{0}\) and \(\vec{b}\) ≠ \(\overrightarrow{0}\). Then,
\((\vec{a} \cdot \vec{b})=|\vec{a}||\vec{b}| \cos \theta\)⇒ \(|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}||\cos \theta|\)
⇒ \(\frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|}=|\cos \theta| \leq 1\)
⇒ \(|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}| .\)
Hence, \(|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}| .\)
Theorem 4 (Triangle Inequality) Prove that \(|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}| \text {. }\)
Proof
When \(\vec{a}\) = \(\overrightarrow{0}\), then |\(\vec{a}\)| = 0.
∴ \(|\vec{a}+\vec{b}|=|\overrightarrow{0}+\vec{b}|=|\vec{b}| \text { and }|\vec{a}|+|\vec{b}|=0+|\vec{b}|=|\vec{b}| \text {. }\)
So, in this case, \(|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|\)
Similarly, when \(\vec{b}\) = \(\overrightarrow{0}\), we have \(|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|\)
Let us consider the case when \(\vec{a}\) ≠ \(\overrightarrow{0}\) and \(\vec{b}\) ≠ \(\overrightarrow{0}\).
Now, \(|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b})^2\)
= \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})\)
= \(\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
= \(|\vec{a}|^2+2(\vec{a} \cdot \vec{b})+|\vec{b}|^2\) [∵ \(\vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}\)]
≤ \(|\vec{a}|^2+2|\vec{a} \cdot \vec{b}|+|\vec{b}|^2 \) [∵ α ≤ |α| ∀ α ∈ R]
≤ \(|\vec{a}|^2+2|\vec{a}| \cdot|\vec{b}|+|\vec{b}|^2\) [∵ \(|\vec{a} \cdot \vec{b}| \leq|\vec{a}| \cdot|\vec{b}|\)]
= \(\{|\vec{a}|+|\vec{b}|\}^2\)
∴ \(|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}| \text {. }\)
Common Questions on Vector Algebra and Their Solutions
Theorem 5 (Distributive Law) Prove that \(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c} .\)
Proof
Let O be the origin, \(\overrightarrow{O A}\) = \(\vec{a}\), \(\overrightarrow{O B}\) = \(\vec{b}\) and \(\overrightarrow{B C}\) = \(\vec{c}\). Then,
\(\overrightarrow{O C}=(\overrightarrow{O B}+\overrightarrow{B C})=(\vec{b}+\vec{c})\)Draw BM ⊥ OA and CN ⊥ OA. Then,
\(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \overrightarrow{O C}\)= \(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \overrightarrow{O C}\), where ∠AOC = θ.
= a x ON [∵ (OC)cos θ = ON]
= a(OM + MN)
= a(OM) + a(MN)
= a(projection of \(\vec{v}\) on \(\vec{a}\)) + a(projection of \(\vec{c}\) on \(\vec{a}\))
= \((\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})\)
Hence, \(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c} .\)
Theorem 6 For any two vectors \(\vec{a}\) and \(\vec{b}\),
(1) \(\vec{a} \cdot(-\vec{b})=-(\vec{a} \cdot \vec{b})=(-\vec{a}) \cdot \vec{b}\)
(2) \((-\vec{a}) \cdot(-\vec{b})=\vec{a} \cdot \vec{b}\)
Proof
Let \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{O B}\) = \(\vec{b}\). Produce AO and BO to A’ and B’ respectively, such that OA’ = OA and OB’ = OB.
Then, \(\overrightarrow{O A^{\prime}}=-\vec{a} \text { and } \overrightarrow{O B^{\prime}}=-\vec{b}\)
Let ∠AOB = θ.
Then, ∠AOB’ = π – θ.
(1) \(\vec{a} \cdot(-\vec{b})=|\vec{a}||-\vec{b}| \cos (\pi-\theta)\)
= \(|\vec{a}||\vec{b}| \cdot(-\cos \theta)\)
= \(-a b \cos \theta=-(\vec{a} \cdot \vec{b}) .\)
∴ \(\vec{a} \cdot(-\vec{b})=-(\vec{a} \cdot \vec{b}) .\)
Similarly, \((-\vec{a}) \cdot \vec{b}=-(\vec{a} \cdot \vec{b}) .\)
Hence, \(\vec{a} \cdot(-\vec{b})=(-\vec{a}) \cdot \vec{b}=-(\vec{a} \cdot \vec{b})\)
(2) \((-\vec{a}) \cdot(-\vec{b})=|-\vec{a}||-\vec{b}| \cos \angle A^{\prime} O B^{\prime}\)
= \(|\vec{a}||\vec{b}| \cos \theta\) [∵ ∠A’OB’ = θ]
= \(a b \cos \theta=\vec{a} \cdot \vec{b} .\)
Hence, \((-\vec{a}) \cdot(-\vec{b})=\vec{a} \cdot \vec{b}\).
Theorem 7 If \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \quad \text { and } \quad \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\), prove that \((\vec{a} \cdot \vec{b})=\left(a_1 b_1+a_2 b_2+a_3 b_3\right) .\)
Proof
We have
\(\vec{a} \cdot \vec{b}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)= \(a_1 \hat{i} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)+a_2 \hat{j} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)+a_3 \hat{k} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)
= \(\left(a_1 b_1\right)(\hat{i} \cdot \hat{i})+\left(a_1 b_2\right)(\hat{i} \cdot \hat{j})+\left(a_1 b_3\right)(\hat{i} \cdot \hat{k})+\left(a_2 b_1\right)(\hat{j} \cdot \hat{i})\)
\(+\left(a_2 b_2\right)(\hat{j} \cdot \hat{j})+\left(a_2 b_3\right)(\hat{j} \cdot \hat{k})+\left(a_3 b_1\right)(\hat{k} \cdot \hat{i})+\left(a_3 b_2\right)(\hat{k} \cdot \hat{j})+\left(a_3 b_3\right)(\hat{k} \cdot \hat{k})\)
= \(\left(a_1 b_1+a_2 b_2+a_3 b_3\right)\)
[∵ \(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1, \hat{i} \cdot \hat{j}=\hat{i} \cdot \hat{k}=\hat{j} \cdot \hat{i}\) = … = 0].
Hence, \(\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)=\left(a_1 b_1+a_2 b_2+a_3 b_3\right)\)
Condition Of Perpendicularity
Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \quad \text { and } \quad \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\)
Then, \(\vec{a}\) ⊥ \(\vec{b}\) ⇔ \(\vec{a}\).\(\vec{b}\) = 0 ⇔ \(a_1 b_1+a_2 b_2+a_3 b_3=0 .\)
Thus, \(\vec{a}\) ⊥ \(\vec{b}\) ⇔ \(a_1 b_1+a_2 b_2+a_3 b_3=0 .\)
Angle Between Two Vectors
Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \quad \text { and } \quad \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\), let θ be the angle between them. Then,
\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)⇔ \(a_1 b_1+a_2 b_2+a_3 b_3=|\vec{a}||\vec{b}| \cos \theta\)
⇔ \(\cos \theta=\frac{a_1 b_1+a_2 b_2+a_3 b_3}{|\vec{a}||\vec{b}|}\)
⇔ \(\theta=\cos ^{-1}\left\{\frac{a_1 b_1+a_2 b_2+a_3 b_3}{\left(\sqrt{a_1^2+a_2^2+a_3^2}\right)\left(\sqrt{b_1^2+b_2^2+b_3^2}\right)}\right\}\)
Solved Examples
Example 1 Let \(\vec{a}\) and \(\vec{b}\) be two given vectors such that \(|\vec{a}|=\sqrt{3},|\vec{b}|=2\) and \(\vec{a} \cdot \vec{b}=\sqrt{6}\). Find the angle between \(\vec{a}\) and \(\vec{b}\).
Solution
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,
\(\vec{a} \cdot \vec{b}=\sqrt{6}\) ⇒ \(|\vec{a}||\vec{b}| \cos \theta=\sqrt{6}\)
⇒ \((\sqrt{3})(2) \cos \theta=\sqrt{6}\)
⇒ \(\cos \theta=\frac{\sqrt{6}}{2 \sqrt{3}}=\frac{1}{\sqrt{2}} ⇒ \theta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)
Hence, the required angle is \(\frac{\pi}{4}\).
Example 2 If \(\vec{a}\) and \(\vec{b}\) are two vectors such that \(|\vec{a}|=|\vec{b}|=\sqrt{2}\) and \(\vec{a}\).\(\vec{b}\) = -1, find the angle between \(\vec{a}\) and \(\vec{b}\).
Solution
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,
\(\vec{a}\).\(\vec{b}\) = -1
⇔ \(|\vec{a}| \cdot|\vec{b}| \cdot \cos \theta=-1\)
⇔ \(\sqrt{2} \times \sqrt{2} \times \cos \theta=-1\)
⇔ \(\cos \theta=\frac{-1}{2}\)
⇔ \(\theta=\frac{2 \pi}{3}\) [∵ 0 ≤ θ ≤ 2π].
Hence, the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{2 \pi}{3}\).
Example 3 Find the angle between the vectors \(\vec{a}=(3 \hat{i}-2 \hat{j}+\hat{k}) \text { and } \vec{b}=(\hat{i}-2 \hat{j}-3 \hat{k}) \text {. }\)
Solution
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\).
Now, \(|\vec{a}|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{14} \text {. }\)
And, \(|\vec{b}|=\sqrt{1^2+(-2)^2+(-3)^2}=\sqrt{14}\)
Now, \(\vec{a} \cdot \vec{b}=(3 \hat{i}-2 \hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+3 \hat{k})\)
= [(3×1)+(-2)(-2)+(1×3)] = (3+4+3) = 10
⇒ \(|\vec{a}||\vec{b}| \cos \theta=10\)
⇒ \((\sqrt{14} \cdot \sqrt{14}) \cos \theta=10\) ⇒ \(\cos \theta=\frac{10}{14}=\frac{5}{7}\)
⇒ \(\theta=\cos ^{-1}\left(\frac{5}{7}\right) .\)
Hence, the required angle is cos-1(5/7).
Applications of Vector Algebra in Real Life
Example 4 If \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k} \text { and } \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}\), find the value of \((\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b}) .\)
Solution
We have
\((\vec{a}+3 \vec{b})=(\hat{i}+\hat{j}+2 \hat{k})+3(3 \hat{i}+2 \hat{j}-\hat{k})\)= \((\vec{a}+3 \vec{b})=(\hat{i}+\hat{j}+2 \hat{k})+3(3 \hat{i}+2 \hat{j}-\hat{k})\)
= \((10 \hat{i}+7 \hat{j}-\hat{k})\), and
\((2 \vec{a}-\vec{b})=2(\hat{i}+\hat{j}+2 \hat{k})-(3 \hat{i}+2 \hat{j}-\hat{k})\)= \((2 \hat{i}+2 \hat{j}+4 \hat{k})-(3 \hat{i}+2 \hat{j}-\hat{k})=(-\hat{i}+5 \hat{k})\)
∴ \((\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b})=(10 \hat{i}+7 \hat{j}-\hat{k}) \cdot(-\hat{i}+0 \hat{j}+5 \hat{k})\)
= [10 x (-1) + 7 x 0 + (-1) x 5] = -15.
Hence, \((\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b})=-15\).
Example 5 Find the value of λ for which the vectors \(\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k} \text { and } \vec{b}=\hat{i}+\lambda \hat{j}-3 \hat{k}\) are perpendicular to each other.
Solution
\(\vec{a}\) ⊥ \(\vec{b}\) ⇔ \(\vec{a}\).\(\vec{b}\) = 0
⇔ \((3 \hat{i}+\hat{j}-2 \hat{k}) \cdot(\hat{i}+\lambda \hat{j}-3 \hat{k})=0\)
⇔ 3 x 1 + 1 x λ + (-2) x (-3) = 0
⇔ λ = -9.
Hence, the required value of λ is -9.
Example 5 If \(\vec{a}=(5 \hat{i}-\hat{j}-3 \hat{k}) \text { and } \vec{b}=(\hat{i}+3 \hat{j}-5 \hat{k})\), then show that \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) are perpendicular to each other.
Solution
We have
\((\vec{a}+\vec{b})=(5 \hat{i}-\hat{j}-3 \hat{k})+(\hat{i}+3 \hat{j}-5 \hat{k})=(6 \hat{i}+2 \hat{j}-8 \hat{k})\), and
\((\vec{a}-\vec{b})=(5 \hat{i}-\hat{j}-3 \hat{k})-(\hat{i}+3 \hat{j}-5 \hat{k})=(4 \hat{i}-4 \hat{j}+2 \hat{k}) .\)∴ \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(6 \hat{i}+2 \hat{j}-8 \hat{k}) \cdot(4 \hat{i}-4 \hat{j}+2 \hat{k})\)
= (24 – 8 – 16) = 0.
Hence, \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) are perpendicular to each other.
Example 7 If \(\overrightarrow{A B}\) = \((2 \hat{i}+\hat{j}-3 \hat{k})\), A(1,2,-1) is the given point then find the coordinates of B.
Solution
Let O be the origin. Then, \(\overrightarrow{O A}\) = \((\hat{i}+2 \hat{j}-\hat{k})\)
∴ \(\overrightarrow{A B}=(\overrightarrow{O B}-\overrightarrow{O A})\)
⇒ \(\overrightarrow{O B}=\overrightarrow{A B}+\overrightarrow{O A}\)
⇒ \(\overrightarrow{O B}=(2 \hat{i}+\hat{j}-3 \hat{k})+(\hat{i}+2 \hat{j}-\hat{k})=(3 \hat{i}+3 \hat{j}-4 \hat{k})\)
Hence, the coordinates of B are (3,3,-4).
Example 8 Find the projection of \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k} \text { on } \vec{b}=\hat{i}-2 \hat{j}+\hat{k}\)
Solution
Projection of \(\vec{a}\) on \(\vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
= \(\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k})}{\sqrt{1^2+(-2)^2+1^2}}=\frac{(2+2+1)}{\sqrt{6}}=\frac{5}{\sqrt{6}} .\)
Example 9 If \(\hat{a}\) is a unit vector such that \((\vec{x}-\hat{a}) \cdot(\vec{x}+\hat{a})\) = 3, then find |\(\vec{x}\)|.
Solution
Since \(\hat{a}\) is a unit vector, we have |\(\hat{a}\)| = 1.
Now, \((\vec{x}-\hat{a}) \cdot(\vec{x}+\hat{a})\) = 3
⇒ \(\vec{x} \cdot \vec{x}+\vec{x} \cdot \hat{a}-\hat{a} \cdot \vec{x}-\hat{a} \cdot \hat{a}=3\)
⇒ \(|\vec{x}|^2+\vec{x} \cdot \hat{a}-\vec{x} \cdot \hat{a}-|\hat{a}|^2=3\) [∵ \(\hat{a}\).\(\vec{x}\) = \(\vec{x}\).\(\hat{a}\)]
⇒ \(|\vec{x}|^2-1=3\) [∵ \(|\hat{a} \cdot|^2=1\)]
⇒ \(|\vec{x}|^2=4\)
⇒ \(|\vec{x}|=2\) [∵ |\(\vec{x}\)| ≥ 0].
Example 10 If \(\vec{a}\) and \(\vec{b}\) are vectors such that |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3 and \(\vec{a}\).\(\vec{b}\) = 4, find \(|\vec{a}-\vec{b}|\).
Solution
We have
\(|\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)= \(\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\) (by distributive law)
= \(|\vec{a}|^2-2 \vec{a} \cdot \vec{b}+|\vec{b}|^2\) [∵ \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)]
= (22 – 2 x 4 + 32) = 5.
Hence, \(|\vec{a}-\vec{b}|\) = √5.
Example 11 Find a vector whose magnitude is 3 units and which is perpendicular to each of the vectors \(\vec{a}=3 \hat{i}+\hat{j}-4 \hat{k} \text { and } \vec{b}=6 \hat{i}+5 \hat{j}-2 \hat{k} \text {. }\)
Solution
Let the required vector be \(\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\)
Then, \(|\vec{c}|=3 \Leftrightarrow \sqrt{c_1^2+c_2^2+c_3^2}=3 \Leftrightarrow c_1^2+c_2^2+c_3^2=9\) …(1)
Also, \(\vec{c}\) ⊥ \(\vec{a}\) ⇒ \(\vec{c}\).\(\vec{a}\) = 0
⇒ \(\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right) \cdot(6 \hat{i}+5 \hat{j}-2 \hat{k})=0\)
⇒ 3c1 + c2 – 4c3 = 0 …(2)
And, \(\vec{c}\) ⊥ \(\vec{b}\) ⇒ \(\vec{c}\).\(\vec{b}\) = 0
⇒ \(\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right) \cdot(6 \hat{i}+5 \hat{j}-2 \hat{k})=0\)
⇒ 6c1 + 5c2 – 2c3 = 0 …(3)
From (2) and (3), by cross multiplication, we get
\(\frac{c_1}{(-2+20)}=\frac{c_2}{(-24+6)}=\frac{c_3}{(15-6)}=k \text { (say) }\)⇒ \(\frac{c_1}{18}=\frac{c_2}{-18}=\frac{c_3}{9}=k \Rightarrow \frac{c_1}{2}=\frac{c_2}{-2}=\frac{c_3}{1}=k\)
⇒ c1 = 2k, c2 = -2 and c3 = 1.
Hence, \(\vec{c}=(2 \hat{i}-2 \hat{j}+\hat{k})\) is the required vector.
Example 12 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) and |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 5, |\(\vec{c}\)| = 7, find the angle between \(\vec{a}\) and \(\vec{b}\).
Solution
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\).
Now, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)
⇒ \((\vec{a}+\vec{b})=-\vec{c}\)
⇒ \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=(-\vec{c}) \cdot(-\vec{c})\)
⇒ \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+2 \vec{a} \cdot \vec{b}=\vec{c} \cdot \vec{c}\) [∵ \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\)]
⇒ \(|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}| \cos \theta=|\vec{c}|^2\)
⇒ 9 + 25 + 2 x 3 x 5 x cos θ = 49 [∵ |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 5, |\(\vec{c}\)| = 7]
⇒ cos θ = \(\frac{1}{2}\)
⇒ θ = 60°.
Hence, the angle between \(\vec{a}\) and \(\vec{b}\) is 60°.
Example 13 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are vectors such that |\(\vec{a}\)| = 5, |\(\vec{b}\)| = 4, |\(\vec{c}\)| = 3 and each is perpendicular to the sum of the other two, find \(|\vec{a}+\vec{b}+\vec{c}|\).
Solution
As given, we have
\(\vec{a} \perp(\vec{b}+\vec{c}), \vec{b} \perp(\vec{c}+\vec{a}) \text { and } \vec{c} \perp(\vec{a}+\vec{b})\)⇒ \(\vec{a} \cdot(\vec{b}+\vec{c})=0, \vec{b} \cdot(\vec{c}+\vec{a})=0 \text { and } \vec{c} \cdot(\vec{a}+\vec{b})=0\)
⇒ \(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=0, \vec{b} \cdot \vec{c}+\vec{b} \cdot \vec{a}=0 \text { and } \vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}=0\) [by the distributive law]
⇒ \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\) …(1)
[using \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\), etc., and adding]
∴ \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})\)
= \((\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})+(\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{b} \cdot \vec{c})+(\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c})\)
= \(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\) [using \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\), etc.]
= (25 + 16 + 9 + 0) = 50
[using (1), and |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and |\(\vec{c}\)| = 5]
⇒ \(|\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}=5 \sqrt{2}\)
Hence, \(|\vec{a}+\vec{b}+\vec{c}|=5 \sqrt{2}\).
Example 14 If \(\hat{a}\) and \(\hat{b}\) are unit vectors and θ is the angle between them, prove that \(\sin \frac{\theta}{2}=\frac{1}{2}|\hat{a}-\hat{b}| .\)
Solution
We have
\(|\hat{a}-\hat{b}|^2=(\hat{a}-\hat{b}) \cdot(\hat{a}-\hat{b})\)= \(\hat{a} \cdot \hat{a}+\hat{b} \cdot \hat{b}-2 \hat{a} \cdot \hat{b}\) [∵ \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\)]
= (1 + 1 – 2 x 1 x 1 x cosθ) [∵ \(\hat{a} \cdot \hat{a}=1 \text { and } \hat{b} \cdot \hat{b}=1\)]
= \(2(1-\cos \theta)=4 \sin ^2 \frac{\theta}{2}\)
∴ \(\sin ^2 \frac{\theta}{2}=\frac{1}{4} \cdot|\hat{a}-\hat{b}|^2 \Rightarrow \sin \frac{\theta}{2}=\frac{1}{2}|\hat{a}-\hat{b}| .\)
Example 15 Let \(\vec{a}\) and \(\vec{b}\) be two nonzero vectors. Prove that \(\vec{a} \perp \vec{b} \Leftrightarrow|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| .\)
Solution
Let \(\vec{a}\) ⊥ \(\vec{b}\). Then, \((\vec{a} \cdot \vec{b})=0\) …(1)
Now, \(|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})\)
= \(\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
= \(|\vec{a}|^2+|\vec{b}|^2\) {∵ \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\) = 0}.
Also, \(|\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)
= \(\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
= \(|\vec{a}|^2+|\vec{b}|^2\) {∵ \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\) = 0}.
Thus, \(|\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2\), and therefore, \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \text {. }\)
∴ \(\vec{a} \perp \vec{b} \Rightarrow|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| .\)
Conversely, suppose that \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| .\) Then,
\(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \Rightarrow|\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2\)⇒ \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)
⇒ \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)
⇒ \(\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
⇒ \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a})=0\)
⇒ \(4(\vec{a} \cdot \vec{b})=0\) [∵ \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\)]
⇒ \(\vec{a}\).\(\vec{b}\) = 0
⇒ \(\vec{a}\) ⊥ \(\vec{b}\).
Thus, \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \Rightarrow \vec{a} \perp \vec{b} .\)
Hence, \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| ⇔ \vec{a} \perp \vec{b} .\)
Example 16 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) then find the value of \((\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\).
Solution
Since \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors, we have
|\(\vec{a}\)| = 1, |\(\vec{b}\)| = 1 and |\(\vec{c}\)| = 1.
Now, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)
⇒ \((\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0\) [∵ \(\overrightarrow{0}\).\(\overrightarrow{0}\) = 0]
⇒ \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\)
⇒ \(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\)
⇒ \((\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-\frac{3}{2}\) [∵ \(|\vec{a}|^2=1,|\vec{b}|^2=1,|\vec{c}|^2=1\)].
Hence, \((\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-\frac{3}{2}\)
Example 17 If \(\vec{a}\).\(\vec{b}\) = \(\vec{a}\).\(\vec{c}\), show that \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\vec{c}\) or \(\vec{a}\) ⊥ \((\vec{b}-\vec{c})\).
Solution
\(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c} \Rightarrow \vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}=0\)⇒ \(\vec{a} \cdot(\vec{b}-\vec{c})=0\)
⇒ \(\vec{a}=\overrightarrow{0} \text { or }(\vec{b}-\vec{c})=\overrightarrow{0} \text { or } \vec{a} \perp(\vec{b}-\vec{c})\)
⇒ \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\vec{c} \text { or } \vec{a} \perp(\vec{b}-\vec{c}) \text {. }\)
Hence, \(\vec{a}\).\(\vec{b}\) = \(\vec{a}\).\(\vec{c}\) ⇒ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{a}\) = \(\vec{c}\) or \(\vec{a}\) ⊥ \((\vec{b}-\vec{c}).\)
Example 18 For any vector \(\vec{a}\) in space, show that \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=\vec{a}\)
Solution
Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\). Then,
\(\vec{a} \cdot \hat{i}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot \hat{i}=a_1(\hat{i} \cdot \hat{i})+a_2(\hat{j} \cdot \hat{i})+a_3(\hat{k} \cdot \hat{i})\)= a1 [∵ \(\hat{i} \cdot \hat{i}=1, \hat{j} \cdot \hat{i}=0 \text { and } \hat{k} \cdot \hat{i}=0\)].
Thus, \(\vec{a}\).\(\hat{i}\) = a1.
Similarly, \(\vec{a}\).\(\hat{i}\) = a2 and \(\vec{a}\).\(\hat{k}\) = a3.
∴ \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)=\vec{a} .\)
Hence, \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=\vec{a} .\)
Example 19 If \(\vec{a}=(2 \hat{i}+2 \hat{j}+3 \hat{k}), \quad \vec{b}=(-\hat{i}+2 \hat{j}+\hat{k}) \text { and } \vec{c}=(3 \hat{i}+\hat{j})\) such that \((\vec{a}+\lambda \vec{b}) \perp \vec{c}\), then find the value of λ.
Solution
We have
\((\vec{a}+\lambda \vec{b})=(2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})\)= \((2-\lambda) \hat{i}+2(1+\lambda) \hat{j}+(3+\lambda) \hat{k}\)
Now,
\((\vec{a}+\lambda \vec{b}) \perp \vec{c} \Rightarrow(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0\)⇒ \(\{(2-\lambda) \hat{i}+2(1+\lambda) \hat{j}+(3+\lambda) \hat{k}\} \cdot(3 \hat{i}+\hat{j}+0 \cdot \hat{k})=0\)
⇒ 3(2-λ) + 2(1+λ) + 0 = 0 ⇒ λ = 8.
Hence, λ = 8.
Example 20 For any two nonzero vectors \(\vec{a}\) and \(\vec{b}\), show that \((a \vec{b}+b \vec{a}) \perp(a \vec{b}-b \vec{a})\)
Solution
We have
\((a \vec{b}+b \vec{a}) \cdot(a \vec{b}-b \vec{a})\)= \(a \vec{b} \cdot a \vec{b}-a \vec{b} \cdot b \vec{a}+b \vec{a} \cdot a \vec{b}-b^2 \vec{a} \cdot \vec{a}\)
= \(a^2(\vec{b} \cdot \vec{b})-a b(\vec{b} \cdot \vec{a})+b a(\vec{a} \cdot \vec{b})-b^2(\vec{a} \cdot \vec{a})\)
= (a2b2 – b2a2) = 0 [∵ ab = ba and (\(\vec{b}\).\(\vec{a}\)) = (\(\vec{a}\),.\(\vec{b}\))
Hence, \((a \vec{b}+b \vec{a}) \perp(a \vec{b}-b \vec{a})\).
Example 21 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three mutually perpendicular vectors of the same magnitude, prove that \((\vec{a}+\vec{b}+\vec{c})\) is equally inclined to the vectors, \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
Solution
\(|\vec{a}|=|\vec{b}|=|\vec{c}|=a\) (say) [given) …(1)
Since \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors, we have
\(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{c}\) = \(\vec{c}\).\(\vec{a}\) = 0 …(2)
Now, \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})\)
= \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)
= \(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\) [using(2)]
= 3a2 [using(1)]
∴ \(\vec{a}+\vec{b}+\vec{c}|=\sqrt{3} a\) …(3)
Suppose \((\vec{a}+\vec{b}+\vec{c})\) makes angles θ1, θ2, θ3 with \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) respectively.
Then, \((\vec{a}+\vec{b}+\vec{c}) \cdot \vec{a}=|\vec{a}+\vec{b}+\vec{c}||\vec{a}| \cos \theta_1\)
= \(\left(\sqrt{3} a \cdot a \cdot \cos \theta_1\right)=\sqrt{3} a^2 \cos \theta_1\)
⇔ \((\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{a}+\vec{c} \cdot \vec{a})=\sqrt{3} a^2 \cos \theta_1\)
⇔ \(|\vec{a}|^2=\sqrt{3} a^2 \cos \theta_1 \Leftrightarrow a^2=\sqrt{3} a^2 \cos \theta_1\)
⇔ \(\cos \theta_1=\frac{1}{\sqrt{3}} \Leftrightarrow \theta_1=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) .\)
Similarly, \(\theta_2=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text { and } \theta_3=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text {. }\)
∴ \(\theta_1=\theta_2=\theta_3=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text {. }\)
Hence, \((\vec{a}+\vec{b}+\vec{c})\) is equally inclined to \(\vec{a}\), \(\vec{b}\), \(\vec{c}\).
Example 22 If A(0,1,1), B(3,1,5) and C(0,3,3) be the vertices of a △ABC, show, using vectors, that △ABC is right angled at C.
Solution
Let O be the origin. Then,
\(\overrightarrow{O A}=0 \hat{i}+\hat{j}+\hat{k}\), \(\overrightarrow{O B}=3 \hat{i}+\hat{j}+5 \hat{k} \text { and } \overrightarrow{O C}=0 \hat{i}+3 \hat{j}+3 \hat{k}\)
∴ \(\overrightarrow{C A}=(\overrightarrow{O A}-\overrightarrow{O C})\)
= \((0 \hat{i}+\hat{j}+\hat{k})-(0 \hat{i}+3 \hat{j}+3 \hat{k})\)
= \((0-0) \hat{i}+(1-3) \hat{j}+(1-3) \hat{k}\)
= \((0 \hat{i}-2 \hat{j}-2 \hat{k})\)
∴ \(\overrightarrow{C B}=(\overrightarrow{O B}-\overrightarrow{O C})\)
= \((3 \hat{i}+\hat{j}+5 \hat{k})-(0 \hat{i}+3 \hat{j}+3 \hat{k})=(3 \hat{i}-2 \hat{j}+2 \hat{k})\)
∴ \(\overrightarrow{C A} \cdot \overrightarrow{C B}=(0 \hat{i}-2 \hat{j}-2 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+2 \hat{k})\)
= [(0.3) + (-2)x(-2) + (-2)x2] = 0
⇒ \(\overrightarrow{C A} \perp \overrightarrow{C B}\)
Hence, △ABC is the right angled at C.
Example 23 Show that the point A(2,-1,1), B(1,-3,-5) and C(3,-4,-4) are the vertices of a right-angled triangle. Also, find the remaining angles of the triangle.
Solution
Let O be the origin. Then,
\(\overrightarrow{O A}=(2 \hat{i}-\hat{j}+\hat{k})\), \(\overrightarrow{O B}=(\hat{i}-3 \hat{j}-5 \hat{k}) \text { and } \overrightarrow{O C}=(3 \hat{i}-4 \hat{j}-4 \hat{k})\)
∴ \(\overrightarrow{A B}=(\overrightarrow{O B}-\overrightarrow{O A})=(-\hat{i}-2 \hat{j}-6 \hat{k})\)
\(\overrightarrow{B C}=(\overrightarrow{O C}-\overrightarrow{O B})=(2 \hat{i}-\hat{j}+\hat{k})\) \(\overrightarrow{C A}=(\overrightarrow{O A}-\overrightarrow{O C})=(-\hat{i}+3 \hat{j}+5 \hat{k})\)∴ \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\)
This shows that A, B, C are the vertices of a triangle.
\(\overrightarrow{B C} \cdot \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(-\hat{i}+3 \hat{j}+5 \hat{k})\)= [2 x (-1) + (-1) x 3 + 1 x 5] = 0.
∴ \(\overrightarrow{B C} \perp \overrightarrow{C A}\) and therefore, ∠C = 90°.
Hence, △ABC is right angled at C.
Now, ∠A us the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\).
∴ \(\overrightarrow{A B} \cdot \overrightarrow{A C}=|\overrightarrow{A B}||\overrightarrow{A C}| \cos A\)
⇒ \(\cos A=\frac{\overrightarrow{A B} \cdot \overrightarrow{A C}}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}\)
= \(\frac{(-\hat{i}-2 \hat{j}-6 \hat{k}) \cdot(\hat{i}-3 \hat{j}-5 \hat{k})}{\left\{\sqrt{(-1)^2+(-2)^2+(-6)^2}\right\} \cdot\left\{\sqrt{1^2+(-3)^2+(-5)^2}\right\}}\)
[∵ \(\overrightarrow{A C}=-\overrightarrow{C A}=(\hat{i}-3 \hat{j}-5 \hat{k})\)]
= \(\frac{\{(-1) \times 1+(-2) \times(-3)+(-6) \times(-5)\}}{\sqrt{41} \times \sqrt{35}}=\sqrt{\frac{35}{41}}\)
⇒ A = \(\cos ^{-1}\left(\sqrt{\frac{35}{41}}\right)\)
Further, ∠B is the angle between \(\overrightarrow{B C}\) and \(\overrightarrow{B A}\).
∴ \(\overrightarrow{B C} \cdot \overrightarrow{B A}=|\overrightarrow{B C}||\overrightarrow{B A}| \cos B\)
⇒ \(\cos B=\frac{\overrightarrow{B C} \cdot \overrightarrow{B A}}{|\overrightarrow{B C}||\overrightarrow{B A}|}\)
= \(\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+6 \hat{k})}{\left\{\sqrt{2^2+(-1)^2+1^2}\right\}\left\{\sqrt{1^2+2^2+6^2}\right\}}\)
[∵ \(\overrightarrow{B A}=-\overrightarrow{A B}=(\hat{i}+2 \hat{j}+6 \hat{k})\)]
= \(\frac{\{2 \times 1+(-1) \times 2+1 \times 6\}}{\sqrt{6} \times \sqrt{41}}=\sqrt{\frac{6}{41}}\)
⇒ B = \(\cos ^{-1}\left(\sqrt{\frac{6}{41}}\right)\)
Hence, A = \(\cos ^{-1}\left(\sqrt{\frac{35}{41}}\right)\) and B = \(\cos ^{-1}\left(\sqrt{\frac{6}{41}}\right)\).
Example 24 Let \((\hat{i}+\hat{j}+\hat{k})\),\((2 \hat{i}+5 \hat{j})\),\((3 \hat{i}+2 \hat{j}-3 \hat{k})\) and \((\hat{i}-6 \hat{j}-\hat{k})\) be the position vectors of points A, B, C, D respectively. Find the angle between AB and CD. Hence, show that AB ∥ CD.
Solution
Let the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) be θ.
Now, \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \((2 \hat{i}+5 \hat{j})-(\hat{i}+\hat{j}+\hat{k})=(\hat{i}+4 \hat{j}-\hat{k})\)
\(\overrightarrow{C D}\) = (p.v. of D) – (p.v. of C)
= \((\hat{i}-6 \hat{j}-\hat{k})-(3 \hat{i}+2 \hat{j}-3 \hat{k})=(-2 \hat{i}-8 \hat{j}+2 \hat{k})\)
∴ \(|\overrightarrow{A B}|=\sqrt{1^2+4^2+(-1)^2}=\sqrt{18}=3 \sqrt{2}\)
\(|\overrightarrow{C D}|=\sqrt{(-2)^2+(-8)^2+2^2}=\sqrt{72}=6 \sqrt{2}\)Now, \(\overrightarrow{A B} \cdot \overrightarrow{C D}=(\hat{i}+4 \hat{j}-\hat{k}) \cdot(-2 \hat{i}-8 \hat{j}+2 \hat{k})=(-2-32-2)=-36\)
∴ \(\cos \theta=\frac{\overrightarrow{A B} \cdot \overrightarrow{C D}}{|\overrightarrow{A B}||\overrightarrow{C D}|}=\frac{-36}{(3 \sqrt{2} \times 6 \sqrt{2})}=\frac{-36}{36}=-1\)
⇒ θ = π.
Hence, AB ∥ CD.
Examples of Scalar and Vector Products
Example 15 Express the vector \(\vec{a}=(5 \hat{i}-2 \hat{j}+5 \hat{k})\) as sum of two vectors such that one is parallel to the vector \(\vec{b}=(3 \hat{i}+\hat{k})\) and the other is perpendicular to \(\vec{b}\).
Solution
Any vector parallel to \(\vec{b}\) is the form λ\(\vec{b}\) for some scalar λ.
Let \(\vec{a}\) = λ\(\vec{b}\) + \(\vec{c}\), where \(\vec{c}\) ⊥ \(\vec{b}\). Then,
\(\vec{c}=(\vec{a}-\lambda \vec{b}) \perp \vec{b}\)⇔ \((\vec{a}-\lambda \vec{b}) \cdot \vec{b}=0 \Leftrightarrow(\vec{a} \cdot \vec{b})-\lambda(\vec{b} \cdot \vec{b})=0\)
⇔ \((5 \hat{i}-2 \hat{j}+5 \hat{k}) \cdot(3 \hat{i}+\hat{k})-\lambda(3 \hat{i}+\hat{k}) \cdot(3 \hat{i}+\hat{k})=0\)
⇔ (15-0+5) – λ(9+1) = 0 ⇔ 10λ = 20 ⇔ λ = 2.
∴ \(\lambda \vec{b}=2 \vec{b}=(6 \hat{i}+2 \hat{k})\)
And, \(\vec{c}=(\vec{a}-2 \vec{b})=(5 \hat{i}-2 \hat{j}+5 \hat{k})-2(3 \hat{i}+\hat{k})=(-\hat{i}-2 \hat{j}+3 \hat{k})\)
Hence, the required vectors are \((6 \hat{i}+2 \hat{k}) \text { and }(-\hat{i}-2 \hat{j}+3 \hat{k}) \text {. }\)
Vector (or Cross) Product of Vectors
Vector Product Of Two Vectors Let \(\vec{a}\) and \(\vec{b}\) be two nonzero, nonparallel vectors, and let θ be the angle between them such that 0 < θ < π.
Then, the vector product of \(\vec{a}\) and \(\vec{b}\) is defined as \(\vec{a} \times \vec{b}=(|\vec{a}||\vec{b}| \sin \theta) \hat{n}\),
where \(\hat{n}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\), such that \(\vec{a}\), \(\vec{b}\), \(\hat{n}\) form a right-handed system.
When a right-handed screw is rotated from \(\vec{a}\) to \(\vec{b}\) and it advances along \(\hat{n}\) then the system is said to be right-handed.
Remark 1 If \(\vec{a}\) and \(\vec{b}\) are parallel or collinear, i.e., when θ = 0 or θ = π then we define, \(\vec{a}\) x \(\vec{b}\) = \(\overrightarrow{0}\).
In particular, \(\vec{a}\) x \(\vec{a}\) = \(\overrightarrow{0}\).
Remark 2 If \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\), we define, \(\vec{a}\) x \(\vec{b}\) = \(\overrightarrow{0}\).
Remark 3 For any vector \(\vec{a}\), we have \(\vec{a} \times \vec{a}=(|\vec{a}||\vec{a}| \sin 0) \hat{n}=0 \hat{n}=\overrightarrow{0}\)
Angle Between Two Vectors Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,
\(\vec{a} \times \vec{b}=(a b \sin \theta) \hat{n} \text {, where }|\vec{a}|=a \text { and }|\vec{b}|=b\)⇒ \(|\vec{a} \times \vec{b}|=a b \sin \theta\)
⇒ \(\sin \theta=\frac{|\vec{a} \times \vec{b}|}{a b}=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\)
⇒ \(\theta=\sin ^{-1}\left\{\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\right\} .\)
Unit Vector Perpendicular To Two Given Vectors A unit vector \(\hat{n}\) perpendicular to each one of \(\vec{a}\) and \(\vec{b}\) is given by
\(\hat{n}=\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}\)Properties of Vector Product
Result 1 Vector product is not commutative.
In fact, we have \((\vec{b} \times \vec{a})=-(\vec{a} \times \vec{b})\)
Proof Let \(\vec{a}\), \(\vec{b}\), \(\hat{n}\) from a right-handed system. Then,
\((\vec{a} \times \vec{b})=(a b \sin \theta) \hat{n}\) …(1)
Then, \(\vec{b}\), \(\vec{a}\) , –\(\hat{n}\) form a right-handed system.
∴ \((\vec{b} \times \vec{a})=(b a \sin \theta)(-\hat{n})\)
⇒ \(-(\vec{b} \times \vec{a})=(a b \sin \theta) \hat{n}\) …(2)
From (1) and (2), we get \((\vec{a} \times \vec{b})=-(\vec{b} \times \vec{a})\)
Result 2 For any two vectors \(\vec{a}\) and \(\vec{b}\), prove that \((-\vec{b}) \times \vec{a}=(\vec{a} \times \vec{b})=\vec{b} \times(-\vec{a})\).
Proof
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\).
Then, the angle between (-\(\vec{b}\)) is \(\vec{a}\) is (π – θ).
And, (-\(\vec{b}\)), \(\vec{a}\), \(\hat{n}\) form a right-handed system.
∴ \((-\vec{b}) \times \vec{a}=|-\vec{b}||\vec{a}| \sin (\pi-\theta) \hat{n}\)
= \((|\vec{b}||\vec{a}| \sin \theta) \hat{n}=(a b \sin \theta) \hat{n}=(\vec{a} \times \vec{b})\)
∴ \((-\vec{b}) \times \vec{a}=(\vec{a} \times \vec{b}) .\)
Similarly, \(\vec{b} \times(-\vec{a})=(\vec{a} \times \vec{b})\)
Hence, \((-\vec{b}) \times \vec{a}=(\vec{a} \times \vec{b})=\vec{b} \times(-\vec{a})\)
Result 3 For any scalar m, we have \((m \vec{a} \times \vec{b})=m(\vec{a} \times \vec{b})=\vec{a} \times(m \vec{b})\)
An Important Note At a later stage, we shall prove that for any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), we have \(\vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c}\), i.e., the position of dot and cross can be interchanged. However, we shall use this fact in proving the following theorem.
Result 4 (Distributive law) For any vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), prove that \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)
Proof
Let \(\vec{p}=[\vec{a} \times(\vec{b}+\vec{c})-(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})]\), and let \(\vec{r}\) be any arbitraty vector.
Then,
\(\vec{r} \cdot \vec{p}=\vec{r} \cdot[\vec{a} \times(\vec{b}+\vec{c})-(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})]\)= \(\vec{r} \cdot[\vec{a} \times(\vec{b}+\vec{c})]-\vec{r} \cdot(\vec{a} \times \vec{b})-\vec{r} \cdot(\vec{a} \times \vec{c})\)
= \((\vec{r} \times \vec{a}) \cdot(\vec{b}+\vec{c})-(\vec{r} \times \vec{a}) \cdot \vec{b}-(\vec{r} \times \vec{a}) \cdot \vec{c}\)
= \((\vec{r} \times \vec{a}) \cdot \vec{b}+(\vec{r} \times \vec{a}) \cdot \vec{c}-(\vec{r} \times \vec{a}) \cdot \vec{b}-(\vec{r} \times \vec{a}) \cdot \vec{c}\)
= 0.
Now, \(\vec{r}\).\(\vec{p}\) = 0 ⇒ \(\vec{r}\) = \(\overrightarrow{0}\) or \(\vec{r}\) = \(\overrightarrow{0}\) or \((\vec{r} ⊥ \vec{p})\).
Since \(\vec{r}\) is an arbitraty vector, we may choose it in such a way that \(\vec{r}\) ≠ \(\overrightarrow{0}\) and \(\vec{r}\) is not perpendicular to \(\vec{p}\),.
Then, \(\vec{p}\) = \(\overrightarrow{0}\)
⇒ \(\vec{a} \times(\vec{b}+\vec{c})-(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})=\overrightarrow{0}\)
⇒ \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)
Hence, \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\).
Result 5 For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), prove that \(\vec{a} \times(\vec{b}-\vec{c})=(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c}) .\)
Proof
We have
\(\vec{a} \times(\vec{b}-\vec{c})=\vec{a} \times[\vec{b}+(-\vec{c})]\)= \((\vec{a} \times \vec{b})+\vec{a} \times(-\vec{c})\) [by the distributive law]
= \((\vec{a} \times \vec{b})+[-(\vec{a} \times \vec{c})]\) [∵ \(\vec{a} \times(-\vec{c})=-(\vec{a} \times \vec{c})\)]
= \((\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})\)
Hence, \(\vec{a} \times(\vec{b}-\vec{c})=(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c}) .\)
Result Prove that two nonzero vectors \(\vec{a}\)and \(\vec{b}\) are parallel or collinear if and only if \((\vec{a} \times \vec{b})=\overrightarrow{0}\)
Proof
Let \(\vec{a}\) ≠ \(\overrightarrow{0}\) and \(\vec{b}\) ≠ \(\overrightarrow{0}\) and let θ be the angle between them.
Let \(\vec{a}\) and \(\vec{b}\) be parallel or collinear.
Then, θ = 0 or θ = π
⇒ sin θ = 0 ⇒ \((\vec{a} \times \vec{b})=(a b \sin \theta) \hat{n}=0 \hat{n}=\overrightarrow{0} .\)
Thus, when \(\vec{a}\) and \(\vec{b}\) are parallel or collinear, then \((\vec{a} \times \vec{b})=\overrightarrow{0}\).
Again, let \(\vec{a}\) ≠ \(\overrightarrow{0}\), \(\vec{b}\) ≠ \(\overrightarrow{0}\) and \((\vec{a} \times \vec{b})=\overrightarrow{0}\). Then
\((\vec{a} \times \vec{b})=\overrightarrow{0}\) ⇒ \(|\vec{a} \times \vec{b}|=0\)
⇒ ab sinθ = 0
⇒ sin θ = 0 [a ≠ 0 and b ≠ 0]
⇒ θ = 0 or θ = π
⇒ (\(vec{a}\) ∥ \(\vec{b}\)) or (\(\vec{a}\) and \(\vec{b}\) are collinear).
Result 7 Prove that
\((\vec{a} \times \vec{b})=\overrightarrow{0} \Rightarrow \vec{a}=\overrightarrow{0} \text {, or } \vec{b}=\overrightarrow{0} \text {, or } \vec{a} \| \vec{b} \text { or } \vec{a} \text { and } \vec{b}\) are collinear.
Proof
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,
\(\vec{a} \times \vec{b}=\overrightarrow{0} \Rightarrow(a b \sin \theta) \hat{n}=\overrightarrow{0}\)⇒ ab sinθ = 0
⇒ a = 0, or b = 0, or sin θ = 0
⇒ \(\vec{a}\) = \(\overrightarrow{0}\), or \(\vec{b}\) = \(\overrightarrow{0}\), or θ = 0 or θ = π
⇒ \(\vec{a}\) = \(\overrightarrow{0}\), or \(\vec{b}\) = \(\overrightarrow{0}\), or \(\vec{a}\) and \(\vec{b}\) are parallel or collinear.
Corollary Show that \(\vec{a}\) x \(\vec{b}\) = \(\vec{a}\) x \(\vec{c}\) does not imply that \(\vec{b}\) = \(\vec{c}\).
Proof
\(\vec{a} \times \vec{b}=\vec{a} \times \vec{c} \Rightarrow \vec{a} \times \vec{b}-\vec{a} \times \vec{c}=\overrightarrow{0}\)⇒ \(\vec{a} \times(\vec{b}-\vec{c})=\overrightarrow{0}\)
⇒ \(\vec{a}=\overrightarrow{0}, \text { or }(\vec{b}-\vec{c})=0 \text {, or } \vec{a} \|(\vec{b}-\vec{c})\)
⇒ \(\vec{a}=\overrightarrow{0}, \text { or }(\vec{b}-\vec{c})=0 \text {, or } \vec{a} \|(\vec{b}-\vec{c})\)
∴ \(\vec{a} \times \vec{b}=\vec{a} \times \vec{c}\) does not always mean that \(\vec{b}\) = \(\vec{c}\).
Summary
(1) \((\vec{a} \times \vec{b})=-(\vec{b} \times \vec{a})\)
(2) \((-\vec{b}) \times \vec{a}=(\vec{a} \times \vec{b})=\vec{b} \times(-\vec{a})\)
(3) \(m \vec{a} \times \vec{b}=m(\vec{a} \times \vec{b})=\vec{a} \times(m \vec{b})\)
(4) \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)
(5) \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)
(6) \(\vec{a} \times \vec{b}=\overrightarrow{0} \Leftrightarrow(\vec{a} \| \vec{b})\) or (\(\vec{a}\) and \(\vec{b}\) are collinear), when \(\vec{a}\) ≠ \(\overrightarrow{0}\) and \(\vec{b}\) ≠ \(\overrightarrow{0}\)
Area of a Parallelogram
Theorem 1 Two adjacent sides of a ∥gm are represented by \(\vec{a}\) and \(\vec{b}\) respectively. Prove that the area of the ∥gm = \(|\vec{a} \times \vec{b}|\).
Proof
Let ABCD be a ∥gm. Join BD.
Let \(\overrightarrow{A B}\) = \(\vec{a}\), \(\overrightarrow{A D}\) = \(\vec{b}\) and ∠BAD = θ.
Draw DL ⊥ AB. Then,
AB = a and DL = AD sin θ = b sin θ.
∴ ar(△ABD) = \(\frac{1}{2}\) x base x altitude
= \(\left(\frac{1}{2} \times A B \times D L\right)=\frac{1}{2} a b \sin \theta=\frac{1}{2}|\vec{a} \times \vec{b}|\)
∴ ar(∥gm ABCD) = 2 x ar(△ABD) = \(|\vec{a} \times \vec{b}|\).
Remark (\(\vec{a}\) x \(\vec{b}\)) is called the vector area of the ∥gm.
Area of a Triangle
Theorem 2 Prove that the area of △ABC, where \(\overrightarrow{A B}\) = \(\vec{a}\) and \(\overrightarrow{A C}\) = \(\vec{b}\), is \(\frac{1}{2}|\vec{a} \times \vec{b}|\).
Proof
In △ABC, let \(\overrightarrow{A B}\) = \(\vec{a}\) and \(\overrightarrow{A C}\) = \(\vec{b}\) and ∠BAC = θ.
Draw CL ⊥ AB. Then,
AB = a and CL = (AC)sin θ = b sin θ
and ar(△ABC) = \(\frac{1}{2}\) x AB x CL
= \(\frac{1}{2} a b \sin \theta=\frac{1}{2}|\vec{a} \times \vec{b}|\)
∴ ar(△ABC) = \(\frac{1}{2}|\vec{a} \times \vec{b}|\).
Remark \(\frac{1}{2}(\vec{a} \times \vec{b})\) is called the vector area of △ABC.
Area of a Quadrilateral
Theorem 3 Prove that the area of a quadrilateral ABCD with diagonals AC and BD is \(\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}|\).
Proof
Vector area of quad. ABCD
= (vector area of △ABC) + (vector area of △ACD)
= \(\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})+\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{A D})\)
= \(-\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{A B})+\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{A D})\)
= \(\frac{1}{2} \cdot\{\overrightarrow{A C} \times(\overrightarrow{A D}-\overrightarrow{A B})\}=\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{B D})\)
∴ ar(quad.ABCD) = \(\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}|\).
Summary
(1) ar(∥gm ABCD) = \(|\vec{a} \times \vec{b}|\), where \(\overrightarrow{A B}\) = \(\vec{a}\) and \(\overrightarrow{A D}\) = \(\vec{b}\).
(2) ar(△ABC) = \(\frac{1}{2}|\vec{a} \times \vec{b}|\), where \(\overrightarrow{A B}\) = \(\vec{a}\) and \(\overrightarrow{A C}\) = \(\vec{b}\).
(3) ar(quad. ABCD) = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B D}|\), where AC and BD are its diagonals.
Vector Product Of An Orthonormal Vector Triad
For mutually perpendicular unit vectors \(\hat{i}, \hat{j}, \hat{k}\) we have
\(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\hat{0}\) \(\hat{i} \times \hat{j}=\hat{k}=-\hat{j} \times \hat{i}, \hat{j} \times \hat{k}=\hat{i}=-\hat{k} \times \hat{j} \text { and } \hat{k} \times \hat{i}=\hat{j}=-\hat{i} \times \hat{k}\)Vector Product in Terms of Components
Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \text { and } \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\). Then,
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|\)
Proof
We have
\(\vec{a} \times \vec{b}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)= \(a_1 b_1(\hat{i} \times \hat{i})+a_1 b_2(\hat{i} \times \hat{j})+a_1 b_3(\hat{i} \times \hat{k})\)
\(+a_2 b_1(\hat{j} \times \hat{i})+a_2 b_2(\hat{j} \times \hat{j})+a_2 b_3(\hat{j} \times \hat{k})\) \(+a_3 b_1(\hat{k} \times \hat{i})+a_3 b_2(\hat{k} \times \hat{j})+a_3 b_3(\hat{k} \times \hat{k})\)= \(\left(a_2 b_3-a_3 b_2\right) \hat{i}+\left(a_3 b_1-a_1 b_3\right) \hat{j}+\left(a_1 b_2-a_2 b_1\right) \hat{k}\)
[∵ \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\hat{0}, \hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i}, \hat{k} \times \hat{i}=\hat{j}\text { and }\)
\(\hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{j}=-\hat{i} \text { and } \hat{i} \times \hat{k}=-\hat{j}\)]
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|\)
Solved Examples
Example 1 If \(\vec{a}=(3 \hat{i}+\hat{j}-4 \hat{k}) \text { and } \vec{b}=(6 \hat{i}+5 \hat{j}-2 \hat{k}) \text {, find }(\vec{a} \times \vec{b}) \text { and }|\vec{a} \times \vec{b}|\)
Solution
We have
\((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -4 \\
6 & 5 & -2
\end{array}\right|\)
= \((-2+20) \hat{i}-(-6+24) \hat{j}+(15-6) \hat{k}\)
= \((18 \hat{i}-18 \hat{j}+9 \hat{k})\)
\(|\vec{a} \times \vec{b}|^2=\left\{(18)^2+(-18)^2+9^2\right\}=729\)⇒ \(|\vec{a} \times \vec{b}|=\sqrt{729}=27 \text {. }\)
Example 2 Find the values of λ and μ for which \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\)
Solution
Let \(\vec{a}=(2 \hat{i}+6 \hat{j}+27 \hat{k}) \text { and } \vec{b}=\hat{i}+\lambda \hat{j}+\mu \hat{k}\). Then,
\((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|\)
= \((6 \mu-27 \lambda) \hat{i}-(2 \mu-27) \hat{j}+(2 \lambda-6) \hat{k} .\)
Now, \((\vec{a} \times \vec{b})=\overrightarrow{0}\)
⇔ 2λ – 6 = 0 and 2μ – 27 = 0
⇔ λ = 3 and μ = \(\frac{27}{2}\).
Hence, λ = 3 and μ = \(\frac{27}{2}\).
Example 3 If \(\vec{a}=(\hat{i}-2 \hat{j}+3 \hat{k}) \text { and } \vec{b}=(2 \hat{i}+3 \hat{j}-5 \hat{k})\), then find \((\vec{a} \times \vec{b})\) and verify that \((\vec{a} \times \vec{b})\) is perpendicular to each one of \(\vec{a}\) and \(\vec{b}\).
Solution
We have
\((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
2 & 3 & -5
\end{array}\right|\)
= \((10-9) \hat{i}-(-5-6) \hat{j}+(3+4) \hat{k}=(\hat{i}+11 \hat{j}+7 \hat{k})\)
Now, \((\vec{a} \times \vec{b}) \cdot \vec{a}=(\hat{i}+11 \hat{j}+7 \hat{k}) \cdot(\hat{i}-2 \hat{j}+3 \hat{k})\)
∴ \((\vec{a} \times \vec{b}) \cdot \vec{a}\) = (1 – 22 + 21) = 0
∴ \((\vec{a} \times \vec{b}) \perp \vec{a} .\)
And, \((\vec{a} \times \vec{b}) \cdot \vec{b}=(\hat{i}+11 \hat{j}+7 \hat{k}) \cdot(2 \hat{i}+3 \hat{j}-5 \hat{k})\) = (2 + 33 – 35) = 0.
∴ \((\vec{a} \times \vec{b}) \perp \vec{b}\)
Example 4 Find a unit vector perpendicular to each one of the vectors \(\vec{a}=(4 \hat{i}-\hat{j}+3 \hat{k}) \text { and } \vec{b}=(2 \hat{i}+2 \hat{j}-\hat{k})\)
Solution
We know that \((\vec{a} \times \vec{b})\) is a vector perpendicular to each one of \(\vec{a}\) and \(\vec{b}\). So, the required vector is \(\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}\).
Now, \((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
4 & -1 & 3 \\
2 & 2 & -1
\end{array}\right|\)
= \((1-6) \hat{i}-(-4-6) \hat{j}+(8+2) \hat{k}\)
= \((-5 \hat{i}+10 \hat{j}+10 \hat{k})\)
\(|\vec{a} \times \vec{b}|=\sqrt{(-5)^2+(10)^2+(10)^2}=\sqrt{225}=15\)Hence, the required unit vector = \(\frac{(-5 \hat{i}+10 \hat{j}+10 \hat{k})}{15}\)
= \(\frac{1}{3}(-\hat{i}+2 \hat{j}+2 \hat{k}) .\)
Example 5 Find a vector of magnitude 15, which is perpendicular to both the vectors \((4 \hat{i}-\hat{j}+8 \hat{k}) \text { and }(-\hat{j}+\hat{k})\)
Solution
Let \(\vec{a}=(4 \hat{i}-\hat{j}+8 \hat{k}) \text { and } \vec{b}=(-\hat{j}+\hat{k})\)
A unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) = \(\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}\).
Now, \(\vec{a} \times \vec{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
4 & -1 & 8 \\
0 & -1 & 1
\end{array}\right|\)
= \((-1+8) \hat{i}-(4-0) \hat{j}+(-4-0) \hat{k}\)
= \((7 \hat{i}-4 \hat{j}-4 \hat{k})\)
∴ \(|\vec{a} \times \vec{b}|=\sqrt{7^2+(-4)^2+(-4)^2}=\sqrt{81}=9 .\)
So, a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\)
= \(|\vec{a} \times \vec{b}|=\sqrt{7^2+(-4)^2+(-4)^2}=\sqrt{81}=9 .\)
The required vector = \(\frac{15(7 \hat{i}-4 \hat{j}-4 \hat{k})}{9}=\frac{5}{3}(7 \hat{i}-4 \hat{j}-4 \hat{k})\)
Example 6 If \(|\vec{a}|=2,|\vec{b}|=7 \text { and }(\vec{a} \times \vec{b})=(3 \hat{i}+2 \hat{j}+6 \hat{k})\), find the angle between \(\vec{a}\) and \(\vec{b}\).
Solution
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,
\(\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}\)⇒ \(|\vec{a} \times \vec{b}|=\sqrt{3^2+2^2+6^2}=\sqrt{49}=7\)
⇒ \(|\vec{a}||\vec{b}| \sin \theta=7\) [∵ \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta\)]
⇒ \(\sin \theta=\frac{7}{|\vec{a}||\vec{b}|}=\frac{7}{(2 \times 7)}=\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{6}\).
Hence, the required angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\).
Example 7 If |\(\vec{a}\)| = √26, |\(\vec{b}\)| = 7 and \(|\vec{a} \times \vec{b}|\) = 35, find \(\vec{a}\).\(\vec{b}\)
Solution
Given that |\(\vec{a}\)| = √26, |\(\vec{b}\)| = 7 and \(|\vec{a} \times \vec{b}|\) = 35.
∴ \(|\vec{a} \times \vec{b}|=35 \Rightarrow|\vec{a}||\vec{b}| \sin \theta=35\)
⇒ \(\sin \theta=\frac{35}{|\vec{a}||\vec{b}|}=\frac{35}{(\sqrt{26}) \times 7}=\frac{5}{\sqrt{26}}\)
Now, \(\cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{25}{26}}=\frac{1}{\sqrt{26}} .\)
∴ \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=\left(\sqrt{26} \times 7 \times \frac{1}{\sqrt{26}}\right)=7 .\)
Hence, \(\vec{a}\).\(\vec{b}\) = 7.
Example 8 If \(\vec{a}=4 \hat{i}+3 \hat{j}+2 \hat{k} \text { and } \vec{b}=3 \hat{i}+2 \hat{k} \text {, find }|\vec{b} \times 2 \vec{a}| \text {. }\)
Solution
We have
\(\vec{b}=(3 \hat{i}+2 \hat{k}) \text { and } 2 \vec{a}=(8 \hat{i}+6 \hat{j}+4 \hat{k}) \text {. }\)∴ \((\vec{b} \times 2 \vec{a})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 0 & 2 \\
8 & 6 & 4
\end{array}\right|\)
= \((0-12) \hat{i}+(16-12) \hat{j}+(18-0) \hat{k}\)
= \((-12 \hat{i}+4 \hat{j}+18 \hat{k})\)
∴ \(|\vec{b} \times 2 \vec{a}|=|-12 \hat{i}+4 \hat{j}+18 \hat{k}|\)
= \(\sqrt{(-12)^2+4^2+(18)^2}=\sqrt{484}=22\)
Hence, \(|\vec{b} \times 2 \vec{a}|\) = 22.
Example 9 Find the sine of the angle between the vectors \(\vec{a}=(2 \hat{i}-\hat{j}+3 \hat{k})\) and \(\vec{b}=(\hat{i}+3 \hat{j}+2 \hat{k})\).
Solution
We have
\((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 3 \\
1 & 3 & 2
\end{array}\right|\)
= \((-2-9) \hat{i}-(4-3) \hat{j}+(6+1) \hat{k}=(-11 \hat{i}-\hat{j}+7 \hat{k})\)
\(|\vec{a} \times \vec{b}|=\sqrt{(-11)^2+(-1)^2+7^2}=\sqrt{171}=3 \sqrt{19}\) \(|\vec{a}|=\sqrt{2^2+(-1)^2+3^2}=\sqrt{14}\) \(|\vec{b}|=\sqrt{1^2+3^2+2^2}=\sqrt{14}\)Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,
\(\sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{3 \sqrt{19}}{(\sqrt{14})(\sqrt{14})}=\frac{3}{14} \sqrt{19} .\)Example 10 Find the area of the parallelogram whose adjacent sides are represented by the vectors \((3 \hat{i}+\hat{j}-2 \hat{k}) \text { and }(\hat{i}-3 \hat{j}+4 \hat{k})\)
Solution
Let \(\vec{a}=(3 \hat{i}+\hat{j}-2 \hat{k}) \text { and } \vec{b}=(\hat{i}-3 \hat{j}+4 \hat{k})\)
Then, vector area of the ∥gm is \((\vec{a} \times \vec{b})\).
Now, \((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -2 \\
1 & -3 & 4
\end{array}\right|\)
= \((4-6) \hat{i}-(12+2) \hat{j}+(-9-1) \hat{k}\)
= \((-2 \hat{i}-14 \hat{j}-10 \hat{k})\)
Required area = \(|\vec{a} \times \vec{b}|\)
= \(\sqrt{(-2)^2+(-14)^2+(-10)^2}\) sq units.
= \(\sqrt{300} \text { sq units }=10 \sqrt{3} \text { sq units. }\)
Example 11 Find the area of the parallelogram whose diagonals are represented by the vectors \(\vec{d}_1=(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \vec{d}_2=(3 \hat{i}+4 \hat{j}-\hat{k}) \text {. }\)
Solution
Given that \(\vec{d}_1=(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \vec{d}_2=(3 \hat{i}+4 \hat{j}-\hat{k}) \text {. }\)
Vector area of the ∥gm is \(\frac{1}{2}\left(\vec{d}_1 \times \vec{d}_2\right)\)
Now, \(\left(\vec{d}_1 \times \vec{d}_2\right)=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & 4 & -1
\end{array}\right|\)
= \((1-4) \hat{i}-(-2-3) \hat{j}+(8+3) \hat{k}\)
= \((-3 \hat{i}+5 \hat{j}+11 \hat{k})\)
Required area = \(\frac{1}{2}\left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right|\)
= \(\frac{1}{2} \sqrt{(-3)^2+5^2+(11)^2}\) sq units
= \(\frac{1}{2} \sqrt{155}\) sq units.
Example 12 Find the area of the triangle whose adjacent sides are determined by the vectors \(\vec{a}=(-2 \hat{i}-5 \hat{k}) \text { and } \vec{b}=(\hat{i}-2 \hat{j}-\hat{k})\)
Solution
Two adjacent sides of the given triangle are represented by the vectors \(\vec{a}=(-2 \hat{i}-5 \hat{k}) \text { and } \vec{b}=(\hat{i}-2 \hat{j}-\hat{k})\)
So, the area of the triangle is \(\frac{1}{2}|\vec{a} \times \vec{b}|\).
Now, \((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 0 & -5 \\
1 & -2 & -1
\end{array}\right|\)
= \((0-10) \hat{i}-(2+5) \hat{j}+(4-0) \hat{k}\)
= \((-10 \hat{i}-7 \hat{j}+4 \hat{k})\)
∴ required area = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
= \(\frac{1}{2} \cdot\left\{\sqrt{(-10)^2+(-7)^2+4^2}\right\}\) sq units
= \(\frac{1}{2} \sqrt{165}\) sq units.
Example 13 Using vector method, find the area of the triangle whose vertices are A(1,1,1), B(1,2,3) and C(2,3,1).
Solution
Position vector of A = \((\hat{i}+\hat{j}+\hat{k})\);
position vector of B = \((\hat{i}+2 \hat{j}+3 \hat{k})\); and
position vector of C = \((2 \hat{i}+3 \hat{j}+\hat{k})\).
∴ \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \((\hat{i}+2 \hat{j}+3 \hat{k})-(\hat{i}+\hat{j}+\hat{k})=(\hat{j}+2 \hat{k}) .\)
And, \(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)
= \((2 \hat{i}+3 \hat{j}+\hat{k})-(\hat{i}+\hat{j}+\hat{k})=(\hat{i}+2 \hat{j})\)
∴ area of △ABC = \(\left|\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})\right|\).
Now, \(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 1 & 2 \\
1 & 2 & 0
\end{array}\right|\)
= \((0-4) \hat{i}+(2-0) \hat{j}+(0-1) \hat{k}=(-4 \hat{i}+2 \hat{j}-\hat{k})\)
⇒ \(\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})=\left(-2 \hat{i}+\hat{j}-\frac{1}{2} \hat{k}\right)\)
⇒ area of △ABC = \(\left|-2 \hat{i}+\hat{j}-\frac{1}{2} \hat{k}\right|\)
= \(\sqrt{(-2)^2+1^2+\left(-\frac{1}{2}\right)^2}=\frac{\sqrt{21}}{2}\) sq units.
Hence, the area of the given triangle is \(\frac{\sqrt{21}}{2}\) sq units.
Example 14 Show that the points whose vectors are \((5 \hat{i}+6 \hat{j}+7 \hat{k}), (7 \hat{i}-8 \hat{j}+9 \hat{k}) \text { and }(3 \hat{i}+20 \hat{j}+5 \hat{k})\) are collinear.
Solution
Let the given points be A, B, C. Then,
\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \((7 \hat{i}-8 \hat{j}+9 \hat{k})-(5 \hat{i}+6 \hat{j}+7 \hat{k})=(2 \hat{i}-14 \hat{j}+2 \hat{k})\).
And, \(\overrightarrow{AC}\) = (p.v. of C) – (p.v. of A)
= \((3 \hat{i}+20 \hat{j}+5 \hat{k})-(5 \hat{i}+6 \hat{j}+7 \hat{k})\)
= \((-2 \hat{i}+14 \hat{j}-2 \hat{k})\).
∴ \(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -14 & 2 \\
-2 & 14 & -2
\end{array}\right|\)
= \(2 \times(-2) \cdot\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 1 \\
1 & -7 & 1
\end{array}\right|=(-4 \times 0)=\overrightarrow{0}\)
[∵ R2 are R3 are identical]
Thus, \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) are parallel vectors, having a common point, A. Hence, the point A, B, C are collinear.
Example 15 Using vector method, show that the points A(2,-1,3), B(4,3,1) and C(3,1,2) are collinear.
Solution
We know, the points A, B, C are collinear ⇔ area of △ABC = 0.
Now, p.v. of A = \((2 \hat{i}-\hat{j}+3 \hat{k})\), p.v. of B = \((4 \hat{i}+3 \hat{j}+\hat{k})\), and p.v. of C = \((3 \hat{i}+\hat{j}+2 \hat{k})\).
\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \(4 \hat{i}+3 \hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})=(2 \hat{i}+4 \hat{j}-2 \hat{k})\)
And, \(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)
= \((3 \hat{i}+\hat{j}+2 \hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})=(\hat{i}+2 \hat{j}-\hat{k})\)
∴ \(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & 4 & -2 \\
1 & 2 & -1
\end{array}\right|=2 \cdot\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -1 \\
1 & 2 & -1
\end{array}\right|=\overrightarrow{0}\)
[∵ R2 and R3 are identical].
Thus, \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) are parallel vectors, having a common point, A. Hence, the points A, B, C are collinear.
Example 15 Show that the points having position vectors \((\vec{a}-2 \vec{b}+3 \vec{c}),(-2 \vec{a}+3 \vec{b}+2 \vec{c}),(-8 \vec{a}+13 \vec{b})\) are collinear, whatever be \(\vec{a}\), \(\vec{b}\), \(\vec{c}\).
Solution
Let A, B, C be the given points whose position vectors are \((\vec{a}-2 \vec{b}+3 \vec{c}),(-2 \vec{a}+3 \vec{b}+2 \vec{c})\) and \((-8 \vec{a}+13 \vec{b})\) respectively.
Then,
\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \((-2 \vec{a}+3 \vec{b}+2 \vec{c})-(\vec{a}-2 \vec{b}+3 \vec{c})=(-3 \vec{a}+5 \vec{b}-\vec{c})\),
and \(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)
= \((-8 \vec{a}+13 \vec{b})-(\vec{a}-2 \vec{b}+3 \vec{c})=(-9 \vec{a}+15 \vec{b}-3 \vec{c})\)
∴ \(\overrightarrow{A B} \times \overrightarrow{A C}=(-3 \vec{a}+5 \vec{b}-\vec{c}) \times(-9 \vec{a}+15 \vec{b}-3 \vec{c})\)
= \(\vec{d} \times 3 \vec{d} \text {, where } \vec{d}=-3 \vec{a}+5 \vec{b}-\vec{c}\)
= \(\overrightarrow{0}\) [∵ \(\vec{d} \times \vec{d}=\overrightarrow{0}\)].
∴ \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) are parallel vectors, a common point, A. Hence, the points A, B, C are collinear.
Lagrange’s Identity
Example 17 Prove that \(|\vec{a} \times \vec{b}|^2=\left|\begin{array}{ll}
\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b}
\end{array}\right| .\)
Solution
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,
\(|\vec{a} \times \vec{b}|^2=(\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{b})\)= \((a b \sin \theta) \hat{n} \cdot(a b \sin \theta) \hat{n}\)
= \(\left(a^2 b^2 \sin ^2 \theta\right)(\hat{n} \cdot \hat{n})=a^2 b^2 \sin ^2 \theta\)
= \(a^2 b^2\left(1-\cos ^2 \theta\right)=a^2 b^2-(a b \cos \theta)^2\)
= \((\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b})-(\vec{a} \cdot \vec{b})^2\)
= \(\left|\begin{array}{ll}
\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b}
\end{array}\right|\)
Hence, \(|\vec{a} \times \vec{b}|^2=\left|\begin{array}{ll}
\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b}
\end{array}\right| .\)
Example 18 Prove that \(\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})=\overrightarrow{0} .\)
Solution
We have
\(\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})\)= \((\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})+(\vec{b} \times \vec{c})+(\vec{b} \times \vec{a})+(\vec{c} \times \vec{a})+(\vec{c} \times \vec{b})\)
[by the distributive law]
= \((\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})-(\vec{a} \times \vec{b})-(\vec{b} \times \vec{c})-(\vec{c} \times \vec{a})\)
= \(\overrightarrow{0}\)
[∵ \((\vec{b} \times \vec{a})=-(\vec{a} \times \vec{b}),(\vec{c} \times \vec{b})=-(\vec{b} \times \vec{c}) \text { and }(\vec{a} \times \vec{c})=-(\vec{c} \times \vec{a})\)].
Hence, \(\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})=\overrightarrow{0} .\)
Example 19 If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), prove that \((\vec{a} \times \vec{b})=(\vec{b} \times \vec{c})=(\vec{c} \times \vec{a})\).
Solution
\(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \Rightarrow \vec{a}+\vec{b}=-\vec{c}\)⇒ \((\vec{a}+\vec{b}) \times \vec{b}=(-\vec{c}) \times \vec{b}\)
⇒ \((\vec{a} \times \vec{b})+(\vec{b} \times \vec{b})=(-\vec{c}) \times \vec{b}\)
[by the distributive law]
⇒ \((\vec{a} \times \vec{b})+\overrightarrow{0}=(\vec{b} \times \vec{c})\)
[∵ \(\vec{b} \times \vec{b}=\overrightarrow{0} \text { and }(-\vec{c}) \times \vec{b}=\vec{b} \times \vec{c}\)]
⇒ \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c}\) …(1)
Also, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \Rightarrow \vec{b}+\vec{c}=-\vec{a}\)
⇒ \((\vec{b}+\vec{c}) \times \vec{c}=(-\vec{a}) \times \vec{c}\)
⇒ \((\vec{b} \times \vec{c})+(\vec{c} \times \vec{c})=(-\vec{a}) \times \vec{c}\) [by the distributive law]
⇒ \((\vec{b} \times \vec{c})+\overrightarrow{0}=\vec{c} \times \vec{a}\)
[∵ \(\vec{c} \times \vec{c}=\overrightarrow{0} \text { and }(-\vec{a}) \times \vec{c}=\vec{c} \times \vec{a}\)]
⇒ \(\vec{b} \times \vec{c}=\vec{c} \times \vec{a}\) …(2)
From (1) and (2), we get \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} .\)
Example 20 Prove that the points A, B, C with position vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are collinear if and only if \((\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})+(\vec{a} \times \vec{b})=\overrightarrow{0}\).
Proof
We have
\(\overrightarrow{A B}\) = (position vector of B) – (position vector of A) = \((\vec{b}-\vec{a})\)
and \(\overrightarrow{B C}\) = (position vector of C) – (position vector of B) = \((\vec{c}-\vec{b})\).
Now, A, B, C are collinear
⇔ \(\overrightarrow{A B}\) and \(\overrightarrow{B C}\) are parallel
⇔ \((\vec{b}-\vec{a}) \times(\vec{c}-\vec{b})=\overrightarrow{0}\)
⇔ \((\vec{b}-\vec{a}) \times \vec{c}-(\vec{b}-\vec{a}) \times \vec{b}=\overrightarrow{0}\) [by the distributive law]
⇔ \(\vec{b} \times \vec{c}-\vec{a} \times \vec{c}-\vec{b} \times \vec{b}+\vec{a} \times \vec{b}=\overrightarrow{0}\) [by the distributive law]
⇔ \((\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})+(\vec{a} \times \vec{b})=\overrightarrow{0}\) [∵ \(\vec{b} \times \vec{b}=\overrightarrow{0} \text { and }-\vec{a} \times \vec{c}=\vec{c} \times \vec{a}\)].
Thus, A, B, C are collinear ⇔ \((\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})+(\vec{a} \times \vec{b})=\overrightarrow{0}\)
Example 21 Prove that \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)
Proof
We have
\((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})\)= \(\vec{a} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{a}-\vec{b} \times \vec{b}\) [by the distributive law]
= \(\vec{a} \times \vec{b}-\vec{b} \times \vec{a}\) [∵ \(\vec{a} \times \vec{a}=\overrightarrow{0} \text { and } \vec{b} \times \vec{b}=\overrightarrow{0}\)]
= \((\vec{a} \times \vec{b})+(\vec{a} \times \vec{b})\) [∵ \(-\vec{b} \times \vec{a}=(\vec{a} \times \vec{b})\)]
= \(2(\vec{a} \times \vec{b})\).
Example 22 If \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a} x \vec{b} = \overrightarrow{0}\), prove that \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\).
Proof
Let \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) x \(\vec{b}\) = \(\overrightarrow{0}\). Then,
\(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) x \(\vec{b}\) = \(\overrightarrow{0}\)
⇒ \((\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text { or } \vec{a} \perp \vec{b}) \text { and }(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text { or } \vec{a} \| \vec{b})\)
⇒ \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}\) [∵ \(\vec{a}\) ⊥ \(\vec{b}\) and \(\vec{a}\) ∥ \(\vec{b}\) can never hold simultaneously].
Hence, \((\vec{a} \cdot \vec{b}=0 \text { and } \vec{a} \times \vec{b}=\overrightarrow{0}) \Rightarrow(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}) \text {. }\)
Example 23 If \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}, \vec{a} \times \vec{b}=\vec{a} \times \vec{c} \text { and } \vec{a} \neq \overrightarrow{0}\), then prove that \(\vec{b}\) = \(\vec{c}.
Proof
[latex]\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c} \text { and } \vec{a} \neq \overrightarrow{0}\)⇒ \(\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}=0 \text { and } \vec{a} \neq \overrightarrow{0}\)
⇒ \(\vec{a} \cdot(\vec{b}-\vec{c})=0 \text { and } \vec{a} \neq \overrightarrow{0}\)
⇒ \(\vec{b}-\vec{c}=\overrightarrow{0} \text { or } \vec{a} \perp(\vec{b}-\vec{c})\)
\(\vec{b}=\vec{c} \text { or } \vec{a} \perp(\vec{b}-\vec{c})\) …(1)
Again, \(\vec{a} \times \vec{b}=\vec{a} \times \vec{c} \text { and } \vec{a} \neq \overrightarrow{0}\)
⇒ \((\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})=\overrightarrow{0} \text { and } \vec{a} \neq \overrightarrow{0}\)
⇒ \(\vec{a} \times(\vec{b}-\vec{c})=\overrightarrow{0} \text { and } \vec{a} \neq \overrightarrow{0}\)
⇒ \((\vec{b}-\vec{c})=\overrightarrow{0} \text { or } \vec{a} \|(\vec{b}-\vec{c})\)
⇒ \(\vec{b}=\vec{c} \text { or } \vec{a} \|(\vec{b}-\vec{c})\) …(2)
From (1) and (2), we get \(\vec{b}\) = \(\vec{c}\)
[∵ \(\vec{a} \perp(\vec{b}-\vec{c}) \text { and } \vec{a} \|(\vec{b}-\vec{c})\) both cannot hold simultaneously].
Example 24 If \(\vec{a} \times \vec{b}=\vec{c} \times \vec{d} \text { and } \vec{a} \times \vec{c}=\vec{b} \times \vec{d}\), show that \((\vec{a}-\vec{d})\) is parallel to \((\vec{b}-\vec{c})\), it being given that a ≠ d and b ≠ c.
Solution
\(\vec{a} \times \vec{b}=\vec{c} \times \vec{d} \text {, and } \vec{a} \times \vec{c}=\vec{b} \times \vec{d}\)⇒ \(\vec{a} \times \vec{b}-\vec{a} \times \vec{c}=\vec{c} \times \vec{d}-\vec{b} \times \vec{d}\)
⇒ \(\vec{a} \times \vec{b}-\vec{a} \times \vec{c}+\vec{b} \times \vec{d}-\vec{c} \times \vec{d}=\overrightarrow{0}\)
⇒ \(\vec{a} \times(\vec{b}-\vec{c})+(\vec{b}-\vec{c}) \times \vec{d}=\overrightarrow{0}\)
⇒ \(\vec{a} \times(\vec{b}-\vec{c})-\vec{d} \times(\vec{b}-\vec{c})=\overrightarrow{0}\)
⇒ \((\vec{a}-\vec{d}) \times(\vec{b}-\vec{c})=\overrightarrow{0}\)
⇒ \((\vec{a}-\vec{d}) \|(\vec{b}-\vec{c})\)
Hence, \((\vec{a}-\vec{d})\) is parallel to \((\vec{b}-\vec{c})\).
Example 25 If \(\vec{a}=\hat{i}+\hat{j}+\hat{k} \text { and } \vec{b}=\hat{j}-\hat{k}\), find a vector \(\vec{c}\) such that \(\vec{a}\) x \(\vec{c}\) = \(\vec{b}\) and \(\vec{a}\).\(\vec{c}\) = 3.
Solution
Let \(\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\). Then,
\((\vec{a} \times \vec{c})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
c_1 & c_2 & c_3
\end{array}\right|\)
= \(\left(c_3-c_2\right) \hat{i}+\left(c_1-c_3\right) \hat{j}+\left(c_2-c_1\right) \hat{k}\)
∴ \((\vec{a} \times \vec{c})=\vec{b} \Rightarrow\left(c_3-c_2\right) \hat{i}+\left(c_1-c_3\right) \hat{j}+\left(c_2-c_1\right) \hat{k}=\hat{j}-\hat{k}\)
⇒ c3 – c2 = 0, c1 – c3 = 1 and c2 – c1 = -1
⇒ c3 = c2, c1 – c3 = 1 and c2 – c1 = -1 …(1)
\(\vec{a} \cdot \vec{c}=(\hat{i}+\hat{j}+\hat{k}) \cdot\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right)\)⇒ \(\vec{a} \cdot \vec{c}=c_1+c_2+c_3\)
⇒ c1 + c2 + c3 = 3 …(2) [∵ \(\vec{a}\).\(\vec{c}\) = 3]
⇒ c1 + c2 + c1 – 1 = 3 [∵ c1 – c3 = 1]
⇒ 2c1 + c2 = 4 …(3)
On solving c1 – c2 = 1 and 2c1 + c2 = 4, we get
3c1 = 5 ⇒ c1 = \(\frac{5}{3}\).
∴ \(c_2=\left(c_1-1\right)=\left(\frac{5}{3}-1\right)=\frac{2}{3} \text {. }\)
And, c3 = c2 = \(\frac{2}{3}\).
Hence, \(\vec{c}=\left(\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right) .\)
Product Of Three Vectors
Scalar Triple Product The scalar triple product of three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) is defined as \((\vec{a} \times \vec{b}) \cdot \vec{c}\), and it is denoted by [\(\vec{a}\), \(\vec{b}\), \(\vec{c}\)].
Thus \([\vec{a} \vec{b} \vec{c}]=(\vec{a} \times \vec{b}) \cdot \vec{c} .\)
Since \((\vec{a} x \vec{b})\) is a vector, \((\vec{a} \times \vec{b}) \cdot \vec{c}\) is a scalar.
Consequently, \([\vec{a} \vec{b} \vec{c}]\) is a scalar quantity.
Some Theorems on Scalar Triple Product
Theorem 1 (Geometrical interpretation) \((\vec{a} \times \vec{b}) \cdot \vec{c}\) represents the volume of a parallelpiped whose coterminous edges are represented by \(\vec{a}\), \(\vec{b}\), \(\vec{c}\).
Proof
Let us consider a parallelepiped having coterminous edges OA, OB and OC respectively.
Let \(\overrightarrow{O A}\) = \(\vec{a}\), \(\overrightarrow{O B}\) = \(\vec{B}\) and \(\overrightarrow{O C}\) = \(\vec{c}\).
Then, \((\vec{a} \times \vec{b})\) is a vector perpendicular to the plane of \(\vec{a}\) and \(vec{b}\).
Let θ be the angle between \((\vec{a} \times \vec{b}) and [latex]vec{c}\).
∴ \([\vec{a} \vec{b} \vec{c}]=(\vec{a} \times \vec{b}) \cdot \vec{c} .\)
= \(|\vec{a} \times \vec{b}||\vec{c}| \cos \theta\)
= (area of ∥gm OADB) [projection of \(\vec{c}\) along \((\vec{a} \times \vec{b})\)]
= (area of ∥gm OADB).OL
= (area of the base of the parallelepiped with coterminous edges \(\vec{a}\), \(\vec{v}\), \(\vec{c}\).
Hence \([\vec{a} \vec{b} \vec{c}]\) represents the volume of the parallelepiped witn coterminous edges \(\vec{A}\), \(\vec{b}\), \(\vec{c}\) forming a right-handed system.
Theorem 2 For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), \((\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})\).
Proof
Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) form a right-handed system, representing the coterminous edges of a parallelepiped of volume V.
Then, V = \((\vec{a} \times \vec{b}) \cdot \vec{c} .\)
Again, \(\vec{b}\), \(\vec{c}\), \(\vec{a}\) form a right-handed system, representing the coterminous edges of the same parallelepiped.
∴ V = \((\vec{b} \times \vec{c}) \cdot \vec{a}=\vec{a} \cdot(\vec{b} \times \vec{c})\) [∵ \(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}\)]
Hence, \((\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})\).
Theorem 3 The scalar triple product of three vectors remains unchanged so long as their cyclic order remains unchanged, i.e., \((\vec{a} \times \vec{b}) \cdot \vec{c}=(\vec{b} \times \vec{c}) \cdot \vec{a}=(\vec{c} \times \vec{a}) \cdot \vec{b} .\)
Proof
Let \(\vec{a}\), \(\vec{b}\), \(\vec{C}\) form a right-handed system, representing the coterminous edges of a rectangular parallelepiped of volume V.
Then, V = \((\vec{a} \times \vec{b}) \cdot \vec{c} .\)
Clearly, \(\vec{b}\), \(\vec{c}\), \(\vec{a}\) as well as \(\vec{c}\), \(\vec{a}\), \(\vec{b}\) form a right-handed system, representing the coterminous edges of the same parallelepiped.
∴ \(V=(\vec{b} \times \vec{c}) \cdot \vec{a} \text { and } V=(\vec{c} \times \vec{a}) \cdot \vec{b} .\)
Thus, \((\vec{a} \times \vec{b}) \cdot \vec{c}=(\vec{b} \times \vec{c}) \cdot \vec{a}=(\vec{c} \times \vec{a}) \cdot \vec{b}\) [each equal to V]
Hence, \([\vec{a} \vec{b} \vec{c}]=[\vec{b} \vec{c} \vec{a}]=[\vec{c} \vec{a} \vec{b}]\)
Theorem 4 The scalar triple product changes in sign but not in magnitude when the cyclic order of vectors is changed.
Proof
For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) we know that
\([\vec{a} \vec{b} \vec{c}]=[\vec{b} \vec{c} \vec{a}]=[\vec{c} \vec{a} \vec{b}]\)∴ \([\vec{c} \vec{b} \vec{a}]=(\vec{c} \times \vec{b}) \cdot \vec{a}=-(\vec{b} \times \vec{c}) \cdot \vec{a}=-[\vec{b} \vec{c} \vec{a}]=-[\vec{a} \vec{b} \vec{c}]\)
And, \([\vec{a} \vec{c} \vec{b}]=(\vec{a} \times \vec{c}) \cdot \vec{b}=-(\vec{c} \times \vec{a}) \cdot \vec{b}=-[\vec{c} \vec{a} \vec{b}]=-[\vec{a} \vec{b} \vec{c}] .\)
∴ \([\vec{c} \vec{b} \vec{a}]=-[\vec{a} \vec{b} \vec{c}] \text { and }[\vec{a} \vec{c} \vec{b}]=-[\vec{a} \vec{b} \vec{c}] \text {. }\)
Theorem 5 The scalar triple product vanishes if any two of its vectors are equal, i.e., \(\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]=0, \left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]=0 \text { and }\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{a}
\end{array}\right]=0.\)
Proof
We have \([\vec{a} \vec{a} \vec{b}]=(\vec{a} \times \vec{a}) \cdot \vec{b}=\overrightarrow{0} \cdot \vec{b}=0\)
\(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]\) [cyclic-ordered property]
= \((\vec{a} \times \vec{a}) \cdot \vec{b}=\overrightarrow{0} \cdot \vec{b}=0 \text {; }\)
and \(\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]\)
= 0. [cyclic-ordered property]
Hence, \(\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{a}
\end{array}\right]=0 .\)
Theorem 6 The scalar triple product vanishes if any two of its vectors are parallel or collinear.
Proof
Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three vectors such that \(\vec{a}\) ∥ \(\vec{b}\), or \(\vec{a}\) and \(\vec{b}\) are collinear.
Then, \(\vec{a} = m\vec{b}\) for some scalar m.
∴ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left[\begin{array}{llll}
m & \vec{b} & \vec{b} & \vec{c}
\end{array}\right]=(m \vec{b} \times \vec{b}) \cdot \vec{c}=\overrightarrow{0} \cdot \vec{c}=0 .\)
Theorem 7 The necessary and sufficient condition for three nonzero, noncollinear vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) to be coplanar is that \([\vec{a} \vec{b} \vec{c}]=0\)
Proof
Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three nonzero, noncollinear and coplanar vectors. Then,
\(\vec{b}\) x \(\vec{c}\) is perpendicular to the plane of \(\vec{b}\) and \(\vec{c}\)
⇒ \((\vec{b} \times \vec{c}) \perp \vec{a}\) [∵ \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar]
⇒ \(\vec{a} \cdot(\vec{b} \times \vec{c})=0\)
⇒ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=0\)
Thus, whenever \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar, we have \([\vec{a} \vec{b} \vec{c}]=0\).
Conversely, let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three nonzero, noncollinear vectors such that \([\vec{a} \vec{b} \vec{c}]=0\). Then,
\([\vec{a} \vec{b} \vec{c}]=0 \Rightarrow \vec{a} \cdot(\vec{b} \times \vec{c})=0\)⇒ \(\vec{a}=\overrightarrow{0}, \text { or }(\vec{b} \times \vec{c})=\overrightarrow{0}, \text { or }(\vec{b} \times \vec{c}) \perp \vec{a}\)
⇒ \((\vec{b} \times \vec{c})=\overrightarrow{0} \text {, or }(\vec{b} \times \vec{c}) \perp \vec{a}\) [∵ \(\vec{a}\) ≠ \(\vec{0}\)]
⇒ \((\vec{b} \times \vec{c}) \perp \vec{a}\)
[∵ \(\vec{b}\) ≠ \(\vec{0}\), \(\vec{c}\) ≠ \(\vec{0}\), and \(\vec{b}\), \(\vec{c}\), are noncollinear ⇒ \(\vec{b} \times \vec{c} \neq \overrightarrow{0}\)].
Thus, \((\vec{b} \times \vec{c})\) is perpendicular to \(\vec{a}\).
But, \((\vec{b} \times \vec{c})\) is perpendicular to the plane of \(\vec{b}\) and \(\vec{c}\).
∴ \(\vec{a}\) must lie in the plane of \(\vec{b}\) and \(\vec{c}\).
Hence \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) must be coplanar.
Theorem 8 (Scalar triple product in terms of components) Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k} \text { and } \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k} \text {. }\)
Then, \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|\)
Proof
We have \(\vec{a} \times \vec{b}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|\)
= \(\left(a_2 b_3-a_3 b_2\right) \hat{i}+\left(a_3 b_1-a_1 b_3\right) \hat{j}+\left(a_1 b_2-a_2 b_1\right) \hat{k}\)
∴ \([\vec{a} \vec{b} \vec{c}]=(\vec{a} \times \vec{b}) \cdot \vec{c}\)
= \(\left(a_2 b_3-a_3 b_2\right) c_1+\left(a_3 b_1-a_1 b_3\right) c_2+\left(a_1 b_2-a_2 b_1\right) c_3\)
= \(a_1\left(b_2 c_3-b_3 c_2\right)+a_2\left(b_3 c_1-b_1 c_3\right)+a_3\left(b_1 c_2-b_2 c_1\right)\)
= \(\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right| .\)
Solved Examples
Example 1 Prove that \([\hat{i} \hat{j} \hat{k}]=1 \text {, and }[\hat{i} \hat{k} \hat{j}]=-1\)
Solution
\([\hat{i} \hat{j} \hat{k}]=(\hat{i} \times \hat{j}) \cdot \hat{k}=\hat{k} \cdot \hat{k}=1\)And, \([\hat{i} \hat{k} \hat{j}]=(\hat{i} \times \hat{k}) \cdot \hat{j}=-\hat{j} \cdot \hat{j}==-1\)
Example 2 If \(\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k} and \vec{c}=-3 \hat{i}+\hat{j}+2 \hat{k} \text {, find }[\vec{a} \vec{b} \vec{c}]\)
Solution
We have
\(\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{rrr}
2 & 1 & 3 \\
-1 & 2 & 1 \\
-3 & 1 & 2
\end{array}\right|\)
= \(\left|\begin{array}{rrr}
0 & 5 & 5 \\
-1 & 2 & 1 \\
0 & -5 & -1
\end{array}\right|\)
= -(-1).[-5+25] = 20.
Example 3 Find the volume of the parallelepiped whose coterminous edges are represented by the vectors \(\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k} \text { and } \vec{c}=2 \hat{i}+\hat{j}-\hat{k}\)
Solution
We have
\(\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{rrr}
2 & -3 & 1 \\
1 & -1 & 2 \\
2 & 1 & -1
\end{array}\right|\)
= \(\left|\begin{array}{rrr}
0 & -1 & -3 \\
1 & -1 & 2 \\
0 & 3 & -5
\end{array}\right|\)
= -(1).(5+9) = -14.
∴ volume of the parallelepiped
= \(|[\vec{a} \vec{b} \vec{c}]|=|-14|\) = 14 cubic units.
Example 4 Show that the vectors \(\hat{i}-3 \hat{j}+4 \hat{k}, 2 \hat{i}-\hat{j}+2 \hat{k} \text { and } 4 \hat{i}-7 \hat{j}+10 \hat{k}\) are coplanar.
Solution
Let \(\vec{a}=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+2 \hat{k} \text { and } \vec{c}=4 \hat{i}-7 \hat{j}+10 \hat{k}\)
∴ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{rrr}
1 & -3 & 4 \\
2 & -1 & 2 \\
4 & -7 & 10
\end{array}\right|=\left|\begin{array}{rrr}
1 & -3 & 4 \\
0 & 5 & -6 \\
0 & 5 & -6
\end{array}\right|\)
= \(\left|\begin{array}{l}
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3-4 R_1
\end{array}\right|\)
= (-30+30) = 0,
Hence, the given vectors are coplanar.
Example 5 Find the value of λ so that the vectors \(\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-3 \hat{k} \text { and } \vec{c}=\hat{j}+\lambda \hat{k}\) are coplanar.
Solution
The given vectors will be coplanar if \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=0\)
Now, \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=0 ⇔ \left|\begin{array}{rrr}
2 & -3 & 1 \\
1 & 2 & -3 \\
0 & 1 & \lambda
\end{array}\right|=0\)
⇔ \(\left|\begin{array}{rrr}
0 & -7 & 7 \\
1 & 2 & -3 \\
0 & 1 & \lambda
\end{array}\right|=0\)
⇔ (-1)(-7λ-7) = 0 ⇔ 7λ + 7 = 0 ⇔ λ = -1.
Hence, the given vectors are coplanar when λ = -1.
Example 6 Show that the four points with position vectors \((4 \hat{i}+5 \hat{j}+\hat{k}),(-\hat{j}-\hat{k}),(3 \hat{i}+9 \hat{j}+4 \hat{k}) \text { and } 4(-\hat{i}+\hat{j}+\hat{k})\) are coplanar.
Solution
Let the given points be A, B, C, D respectively.
Points A, B, C, D are coplanar ⇔ \(\overrightarrow{A B}, \overrightarrow{A C} \text { and } \overrightarrow{A D}\) are coplanar
⇔ \([\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=0\)
Now, \(\overrightarrow{A B}\) = (p.v. of B)-(p.v. of A)
= \((-\hat{j}-\hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=(-4 \hat{i}-6 \hat{j}-2 \hat{k}) .\)
\(\overrightarrow{A C}\) = (p.v. of C)-(p.v. of A)
= \((3 \hat{i}+9 \hat{j}+4 \hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=(-\hat{i}+4 \hat{j}+3 \hat{k})\)
\(\overrightarrow{A D}\) = (p.v. of D)-(p.v. of A)
= \((-4 \hat{i}+4 \hat{j}+4 \hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=(-8 \hat{i}-\hat{j}+3 \hat{k})\)
∴ \(\left[\begin{array}{lll}
\overrightarrow{A B} & \overrightarrow{A C} & \overrightarrow{A D}
\end{array}\right]=\left|\begin{array}{rrr}
-4 & -6 & -2 \\
-1 & 4 & 3 \\
-8 & -1 & 3
\end{array}\right|\)
= \(\left|\begin{array}{rrr}
0 & -22 & -14 \\
-1 & 4 & 3 \\
0 & -21 & -33
\end{array}\right|\)
R_1 \rightarrow R_1-4 R_2 \\
R_3 \rightarrow R_3-8 R_2
\end{array}\right\}\)
= -(-1)[462-462] = 0.
∴ \(\overrightarrow{A B}\), \(\overrightarrow{A C}\) and \(\overrightarrow{A D}\) are coplanar.
Hence, the points A, B, C, D are coplanar.
Example 7 Find the value of λ so that the four points with position vectors \((-6 \hat{i}+3 \hat{j}+2 \hat{k}),(3 \hat{i}+\lambda \hat{j}+4 \hat{k}),(5 \hat{i}+7 \hat{j}+3 \hat{k})\) and \((-13 \hat{i}+17 \hat{j}-2 \hat{k})\) are coplanar.
Solution
Let the given points be A, B, C, D respectively. Then,
\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \((3 \hat{i}+\lambda \hat{j}+4 \hat{k})-(-6 \hat{i}+3 \hat{j}+2 \hat{k})=9 \hat{i}+(\lambda-3) \hat{j}+2 \hat{k}\)
\(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)
= \((5 \hat{i}+7 \hat{j}+3 \hat{k})-(-6 \hat{i}+3 \hat{j}+2 \hat{k})=(11 \hat{i}+4 \hat{j}+\hat{k})\)
\(\overrightarrow{A D}\) = (p.v. of D) – (p.v. of A)
= \((-13 \hat{i}+17 \hat{j}-\hat{k})-(-6 \hat{i}+3 \hat{j}+2 \hat{k})=(-7 \hat{i}+14 \hat{j}-3 \hat{k})\)
Now, A, B, C, D are coplanar
⇔ \(\left[\begin{array}{lll}
\overrightarrow{A B} & \overrightarrow{A C} & \overrightarrow{A D}
\end{array}\right]=0\)
⇔ \(\left|\begin{array}{rcr}
9 & \lambda-3 & 2 \\
11 & 4 & 1 \\
-7 & 14 & -3
\end{array}\right|=0\)
⇔ 9(-12-14) – (λ-3)(-33+7) + 2(154+28) = 0
⇔ -234 + 26λ – 78 + 364 = 0
⇔ 26λ = -52 ⇔ λ = -2.
Hence, the required value of λ is -2.
Example 8 Show that the points A(-1,4,-3), B(3,2,-5), C(-3,8,-5) and D(-3,2,1) are coplanar.
Solution
Clearly, the position vectors of A, B, C, D are \((-\hat{i}+4 \hat{j}-3 \hat{k}), (3 \hat{i}+2 \hat{j}-5 \hat{k}),(-3 \hat{i}+8 \hat{j}-5 \hat{k}) and (-3 \hat{i}+2 \hat{j}+\hat{k})\) respectively.
∴ \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)
= \((3 \hat{i}+2 \hat{j}-5 \hat{k})-(-\hat{i}+4 \hat{j}-3 \hat{k})=(4 \hat{i}-2 \hat{j}-2 \hat{k})\)
\(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)
= \((-3 \hat{i}+8 \hat{j}-5 \hat{k})-(-\hat{i}+4 \hat{j}-3 \hat{k})=(-2 \hat{i}+4 \hat{j}-2 \hat{k})\)
\(\overrightarrow{A D}\) = (p.v. of D) – (p.v. of A)
= \((-3 \hat{i}+2 \hat{j}+\hat{k})-(-\hat{i}+4 \hat{j}-3 \hat{k})=(-2 \hat{i}-2 \hat{j}+4 \hat{k})\)
∴ \(\left[\begin{array}{lll}
\overrightarrow{A B} & \overrightarrow{A C} & \overrightarrow{A D}
\end{array}\right]=\left|\begin{array}{rrr}
4 & -2 & -2 \\
-2 & 4 & -2 \\
-2 & -2 & 4
\end{array}\right|\)
= \(2 \times(-2) \times(-2) \cdot\left|\begin{array}{rrr}
2 & -1 & -1 \\
1 & -2 & 1 \\
1 & 1 & -2
\end{array}\right|\)
= \(8 \cdot\left|\begin{array}{rrr}
0 & -3 & 3 \\
0 & -3 & 3 \\
1 & 1 & -2
\end{array}\right|\)
= (8 x 0) = 0 [∵ R1 and R2 are identical]
⇒ \(\overrightarrow{A B}\), \(\overrightarrow{A C}\), \(\overrightarrow{A D}\) are coplanar
⇒ the points A, B, C, D are coplanar.
Example 9 For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) show that the vectors \((\vec{a}-\vec{b}),(\vec{b}-\vec{c}),(\vec{c}-\vec{a})\) are coplanar.
Solution
We know that \((\vec{a}-\vec{b}),(\vec{b}-\vec{c}),(\vec{c}-\vec{a})\) are coplanar
⇔ \(\left[\begin{array}{lll}
\vec{a}-\vec{b} & \vec{b}-\vec{c} & \vec{c}-\vec{a}
\end{array}\right]=0 .\)
Now, \(\left[\begin{array}{lll}
\vec{a}-\vec{b} & \vec{b}-\vec{c} & \vec{c}-\vec{a}
\end{array}\right]\)
= \((\vec{a}-\vec{b}) \cdot[(\vec{b}-\vec{c}) \times(\vec{c}-\vec{a})]\)
= \((\vec{a}-\vec{b}) \cdot[\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{c} \times \vec{c}+\vec{c} \times \vec{a}]\) [by distribution law]
= \((\vec{a}-\vec{b}) \cdot[\vec{b} \times \vec{c}+\vec{a} \times \vec{b}+\vec{c} \times \vec{a}] [∵ -\vec{b} \times \vec{a}=\vec{a} \times \vec{b} \text {, and } \vec{c} \times \vec{c}=\overrightarrow{0}]\)
= \(\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{a} \times \vec{b})+\vec{a} \cdot[\vec{c} \times \vec{a}]-\vec{b} \cdot(\vec{b} \times \vec{c})-\vec{b} \cdot(\vec{a} \times \vec{b})-\vec{b} \cdot(\vec{c} \times \vec{a})\)
= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{a}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{b} & \vec{c}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\)
= \([\vec{a} \vec{b} \vec{c}]-[\vec{b} \vec{c} \vec{a}]\)
[∵ scalar triple product with two equal vectors is 0]
= \([\vec{a} \vec{b} \vec{c}]-[\vec{a} \vec{b} \vec{c}]=0\)
{∵ \(\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)}.
Hence, \((\vec{a}-\vec{b}),(\vec{b}-\vec{c}),(\vec{c}-\vec{a})\) are coplanar.
Example 10 Prove that \(\left[\begin{array}{llll}
\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}
\end{array}\right]=2\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] .\)
Solution
We have
\([\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]\)= \((\vec{a}+\vec{b}) \cdot[(\vec{b}+\vec{c}) \times(\vec{c}+\vec{a})]\)
= \((\vec{a}+\vec{b}) \cdot[\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{c}+\vec{c} \times \vec{a}]\) [by distribution law]
= \((\vec{a}+\vec{b}) \cdot[(\vec{b} \times \vec{c})-(\vec{a} \times \vec{b})+(\vec{c} \times \vec{a})]\)
[∵ \(\vec{c} \times \vec{c}=\overrightarrow{0}, \text { and } \vec{b} \times \vec{a}=-\vec{a} \times \vec{b}\)]
= \(\vec{a} \cdot(\vec{b} \times \vec{c})-\vec{a} \cdot(\vec{a} \times \vec{b})+\vec{a} \cdot(\vec{c} \times \vec{a})+\vec{b} \cdot(\vec{b} \times \vec{c})-\vec{b} \cdot(\vec{a} \times \vec{b})+\vec{b} \cdot(\vec{c} \times \vec{a})\)
[by distribution law]
= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]-\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{b} & \vec{c}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{b}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\)
= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\)
[∵ scalar triple product with two equal vectors is 0]
= \(2[\vec{a} \vec{b} \vec{c}]\) {∵ \(\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)}
Hence, \(\left[\begin{array}{lll}
\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}
\end{array}\right]=2\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] .\)
Example 11 Show that \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are noncoplanar if and only if \(\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}\) are noncoplanar.
Solution
\(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are noncoplanar
⇔ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] \neq 0 ⇔2 \left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] \neq 0\)
⇔ \(\left[\begin{array}{lll}
\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}
\end{array}\right] \neq 0\)
⇔ \(\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}\) are noncoplanar.
Example 12 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are the position vectors of points A, B, C prove that \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\) is a vector perpendicular tot he plane of triangle ABC.
Solution
In order to prove the required result, we have to show that \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\) is perpendicular to each of the vectors \(\overrightarrow{A B}\), \(\overrightarrow{B C}\) and \(\overrightarrow{C A}\).
We have, \(\overrightarrow{A B}=(\vec{b}-\vec{a}), \overrightarrow{B C}=(\vec{c}-\vec{b}) \text { and } \overrightarrow{C A}=(\vec{a}-\vec{c}) \text {. }\)
Now, \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}) \cdot(\vec{b}-\vec{a})\)
= \((\vec{a} \times \vec{b}) \cdot(\vec{b}-\vec{a})+(\vec{b} \times \vec{c}) \cdot(\vec{b}-\vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{b}-\vec{a})\)
= \((\vec{a} \times \vec{b}) \cdot \vec{b}-(\vec{a} \times \vec{b}) \cdot \vec{a}+(\vec{b} \times \vec{c}) \cdot \vec{b}-(\vec{b} \times \vec{c}) \cdot \vec{a}+(\vec{c} \times \vec{a}) \cdot \vec{b}-(\vec{c} \times \vec{a}) \cdot \vec{a}\)
= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{c} & \vec{a} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{c} & \vec{a} & \vec{a}
\end{array}\right]\)
= \(\left[\begin{array}{lll}
\vec{c} & \vec{a} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\)
[∵ scalar triple product with two equal vectors is 0].
= 0 (∵ \(\left[\begin{array}{lll}
\vec{c} & \vec{a} & \vec{b}
\end{array}\right]=\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\))
Similarly, \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}) \cdot(\vec{c}-\vec{b})=0\)
And, \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}) \cdot(\vec{a}-\vec{c})=0\)
Thus, \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\) is perpendicular to each one of the vectors \(\overrightarrow{A B}\), \(\overrightarrow{B C}\) and \(\overrightarrow{C A}\), and therefore, it is perpendicular to the plane of △ABC.
Example 13 Show that \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar if and only if \((\vec{a}+\vec{b}),(\vec{b}+\vec{c}),(\vec{c}+\vec{a})\) are coplanar.
Solution \((\vec{a}+\vec{b}),(\vec{b}+\vec{c}),(\vec{c}+\vec{a})\) are coplanar
⇔ \(\left[\begin{array}{lll}
\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}
\end{array}\right]=0\)
⇔ \((\vec{a}+\vec{b}) \cdot\{(\vec{b}+\vec{c}) \times(\vec{c}+\vec{a})\}=0\)
⇔ \((\vec{a}+\vec{b}) \cdot\{\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{c}+\vec{c} \times \vec{a}\}=0\)
⇔ \((\vec{a}+\vec{b}) \cdot\{\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{a}\}=0\)
[∵ \(\vec{c} \times \vec{c}=0\)]
⇔ \(\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{b} \times \vec{a})+\vec{a} \cdot(\vec{c} \times \vec{a})\)
\(+\vec{b} \cdot(\vec{b} \times \vec{c})+\vec{b} \cdot(\vec{b} \times \vec{a})+\vec{b} \cdot(\vec{c} \times \vec{a})=0\)⇔ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{b} & \vec{c}
\end{array}\right]\)
\vec{b} & \vec{b} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=0\)
⇔ \([\vec{a} \vec{b} \vec{c}]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=0\)
[∵ scalar triple product with two equal vectors is 0]
⇔ \(2[\vec{a} \vec{b} \vec{c}]=0\) {∵ \(\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)}
⇔ \([\vec{a} \vec{b} \vec{c}]=0\)
⇔ \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.
Hence, \((\vec{a}+\vec{b}), \quad(\vec{b}+\vec{c}), \quad(\vec{c}+\vec{a})\) are coplanar ⇔ \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.
Example 14. For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) prove that \(\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{a}+\vec{b}+\vec{c}
\end{array}\right]=0\)
Solution
We have
\(\left[\begin{array}{lll}\vec{a} & \vec{b}+\vec{c} & \vec{a}+\vec{b}+\vec{c}
\end{array}\right]\)
= \(\{\vec{a} \times(\vec{b}+\vec{c})\} \cdot(\vec{a}+\vec{b}+\vec{c})\)
= \(\{(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\} \cdot(\vec{a}+\vec{b}+\vec{c})\) [by the distribution law]
= \((\vec{a} \times \vec{b}) \cdot \vec{a}+(\vec{a} \times \vec{b}) \cdot \vec{b}+(\vec{a} \times \vec{b}) \cdot \vec{c}+(\vec{a} \times \vec{c}) \cdot \vec{a}\)
\(+(\vec{a} \times \vec{c}) \cdot \vec{b}+(\vec{a} \times \vec{c}) \cdot \vec{c}\) [by the distribution law]
= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{b}
\end{array}\right]\)
[∵ scalar triple product with two equal vectors is 0]
= \([\vec{a} \vec{b} \vec{c}]-[\vec{a} \vec{b} \vec{c}]=0\)
Hence, \(\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{a}+\vec{b}+\vec{c}
\end{array}\right]=0\)