## Three Dimensional Resolution Of Vectors Algebraic Representation

Let the magnitude and direction of a vector \(\vec{r}\), in space, be represented by \(\overrightarrow{O P}\). Our aim here is to find an elegant representation of \(\vec{r}\)

O is taken as the origin of the 3-dimensional space and three mutually perpendicular axes x, y, and z are drawn. A special rule is followed to indicate the directions of the three axes.

If we hold the first 3 fingers of our right hand at right angles to one another, then the forefinger points toward the x-axis, the middle finger toward the y-axis, and the thumb towards the z-axis. This is called the right-handed cartesian coordinate system.

Here, the origin O (0,0,0) is the initial point, and P (x, y, z) is the endpoint of the vector \(\overrightarrow{O P}\). Taking the line segment OP as a diagonal, the cuboid ADPEOBFC is drawn.

The projections of the vector \(\vec{r}\) along the three axes are, from OA = x, OB = AD = y, and OC = DP = z. These x, y, and z are three mutually perpendicular components of the vector \(\vec{r}\).

When the line segments OA, AD, and DP are taken as vectors, the three-dimensional polygon OADP gives, as per the law of polygon of vector addition, \(\overrightarrow{O A}+\overrightarrow{A D}+\overrightarrow{D P}=\overrightarrow{O P}\)….(1)

If \(\hat{i}, \hat{j}, \hat{k}\) are the unit vectors along the x, y and z axes respectively in the positive direction,

∴ \(\overrightarrow{O A}=x \hat{i}, \overrightarrow{A D}=y \hat{j} \quad \text { and } \overrightarrow{D P}=z \hat{k}\)

Since, \(\vec{r}=\overrightarrow{O P}\), from equation (1)

∴ \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)………..(2)

⇒ \(\vec{r}\) is the position vector of the point P w.r.t origin O.

Equation (2) shows the algebraic representation of a position vector with initial point (0,0,0) and final point (x, y, Z).

**Determination Of The Magnitude Of The Vector:** From the geometrical property of a cuboid, (diagonal)² = (length)³ + (breadth)2 + (height)²

or, OP² = OA² + OB² + OC² or, r² = x² + y² + z²

or, \(r=\sqrt{x^2+y^2+z^2}\)….(3)

Hence, the magnitude or value of a vector with its initial point as the origin can be determined easily with the help of the coordinates of its terminal point.

**Direction Cosine:** Let \(\overrightarrow{O P}\) be inclined at an angle a with the x-axis, i.e., ∠POA = α. As per the property of the cuboid, ΔOAP is a right-angled triangle with ∠OAP = 90°.

Hence, \(\cos \alpha=\frac{O A}{O P}=\frac{x}{r}=l\) (say)….(4)

Similarly, if \(\overrightarrow{O P}\) makes angles β and γ with y and z axes respectively,

⇒ \(\left.\begin{array}{rl}\cos \beta =\frac{y}{r}=m \\ \text { and } \cos \gamma =\frac{z}{r}=n\end{array}\right\}\)

From (4) and (5), \(l^2+m^2+n^2=\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=\frac{x^2}{r^2}+\frac{y^2}{r^2}+\frac{z^2}{r^2}\)

= \(\frac{x^2+y^2+z^2}{r^2}=\frac{r^2}{r^2}=1\)…..(6) [with the help of equation (3)]

Hence, to know the direction of the vector, the angles α, β,γ can be determined using equations (4) and (5). cos α, cosβ, cosγ or l, m, n is called the direction cosines of vector r with respect to the axes x, y, and z respectively. Equation (6) indicates the relationship among the direction cosines.

**Sum And Difference Of Two Vectors**: Let \(\overrightarrow{r_1}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text { and } \vec{r}_2=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\) be two vectors.

Sum or resultant of these two vectors is \(\vec{r}_1+\vec{r}_2=\left(x_1+x_2\right) \hat{i}+\left(y_1+y_2\right) \hat{j}+\left(z_1+z_2\right) \hat{k}\)

and its magnitude is, \(\left|\vec{r}_1+\vec{r}_2\right|=\sqrt{\left(x_1+x_2\right)^2+\left(y_1+y_2\right)^2+\left(z_1+z_2\right)^2}\)

The direction of the resultant is specified by the direction cosines. They are \(\cos \alpha=\frac{x_1+x_2}{\left|\vec{r}_1+\vec{r}_2\right|}, \cos \beta=\frac{y_1+y_2}{\left|\vec{r}_1+\vec{r}_2\right|} \text { and } \cos \gamma=\frac{z_1+z_2}{\left|\vec{r}_1+\vec{r}_2\right|}\)

Similarly, the difference between the two vectors, \(\overrightarrow{r_1} \text { and } \overrightarrow{r_2}\) can also be found out. The resultant in this case will be \(\vec{r}_2-\vec{r}_1=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)

The magnitude of the resultant will be, \(\left|\vec{r}_2-\vec{r}_1\right|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

The corresponding direction cosines are \(\cos \alpha=\frac{x_2-x_1}{\left|\overrightarrow{r_2}-\vec{r}_1\right|}, \cos \beta=\frac{y_2-y_1}{\left|\vec{r}_2-\vec{r}_1\right|} \text { and } \cos \gamma=\frac{z_2-z_1}{\left|\vec{r}_2-\vec{r}_1\right|}\)

These relations are applied to determine the sum or difference of any number of vectors.

**Sum And Difference Of Two Vectors Discussions:**

1. Equation (2) represents the 3-dimensional cartesian form of any vector \(\vec{r}\). However, the symbol \(\vec{r}\) is usually reserved for the position vector of a particle.

Any vector \(\vec{A}\) is represented in the system as \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\)….(7)

where A_{x}, A_{y}, and A_{z} stand for the x, y, and z components respectively of the vector \(\vec{A}\). Then, the magnitude of \(\vec{A}\) is

A= \(\sqrt{A_x^2+A_y^2+A_z^2}\)….(8)

The direction cosines, respectively, are l = \(\frac{A_x}{A}, m=\frac{A_y}{A} \quad \text { and } n=\frac{A_z}{A}\)…..(9)

with \(l^2+m^2+n^2=1\)….(10)

2. This is an example of the resolution of a vector \(\vec{A}\) into three components \(A_x \hat{i}, A_y \hat{j} \text { and } A_z \hat{k}\).

We dealt with coplanar vectors, and it was sufficient to resolve a vector into two components only. But, for a system of non-coplanar vectors, it is absolutely necessary to resolve each vector into three mutually perpendicular components.

3. Once every vector under consideration in a problem can be represented in the form of equation (7), vector geometry essentially transforms into vector algebra, and geometrical figures and rules are no longer necessary. Whereas vector geometry is practicable only for a few vectors, there is no limit on the number of vectors that can be handled with the help of vector algebra.

## Algebraic Representation Numerical Examples

**Example 1. The coordinates of the endpoint of a vector \(\overrightarrow{O P}\) is (4,3, -5). Express the vector in terms of its coordinates and find its absolute value.**

**Solution:**

If the coordinates of P is (x, y, z) then, \(\overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k}\)

Here, x=4, y=3 and z=-5

∴ \(\overrightarrow{O P}=4 \hat{i}+3 \hat{j}-5 \hat{k}\)

and the absolute value of the vector,

OP = \(|\overrightarrow{O P}|=\sqrt{x^2+y^2+z^2}\)

= \(\sqrt{4^2+3^2+(-5)^2}\)

= \(\sqrt{16+9+25}=\sqrt{50}=5 \sqrt{2}\)

**Example 2. Find the magnitude of the vector \(\vec{A}=\hat{i}-2 \hat{j}+3 \hat{k}\). Also, find the unit vector in the direction of \(\vec{A}\)**

**Solution:**

Magnitude of \(\vec{A}\), A = \(\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}\)

Unit vector in the direction of \(\vec{A}\) is \(\hat{n}=\frac{\vec{A}}{\vec{A}}=\frac{1}{\sqrt{14}} \hat{i}-\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\)

**Example 3. Vectors \(\vec{A}\) and \(\vec{B}\) can be expressed as \(\vec{A}=10 \hat{i}-12 \hat{j}+5 \hat{k} \text { and } \vec{B}=7 \hat{i}+8 \hat{j}-12 \hat{k}\), where \(\hat{i}\), \(\hat{j}\), k\(\hat{k}\) are unit vectors along x, y, z axes respectively. Find the resultant of the two vectors and its magnitude.**

**Solution:**

The resultant vector \(\vec{C}=\vec{A}+\vec{B}=(10+7) \hat{i}+(-12+8) \hat{j}+(5-12) \hat{k}\)

or, \(\vec{C}=17 \hat{i}-4 \hat{j}-7 \hat{k}\)

Magnitude of the resultant, C = \(\sqrt{17^2+(-4)^2+(-7)^2}=\sqrt{354}=18.81 .\)

**Example 4. Position coordinates of A and B are (-1,5,7) and (3,2,-5) respectively. Express \(\overrightarrow{A B}\) in terms of position coordinates.**

**Solution:**

Let O be the origin, \(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}\) and \(\overrightarrow{A B}=\vec{c}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{b}-\vec{a}\)

Here, \(\vec{a}=-\hat{i}+5 \hat{j}+7 \hat{k} and \vec{b}=3 \hat{i}+2 \hat{j}+(-5) \hat{k}\)

∴ \(\overrightarrow{A B}=\vec{c}=\vec{b}-\vec{a}\)

= \((3 \hat{i}+2 \hat{j}-5 \hat{k})-(-\hat{i}+5 \hat{j}+7 \hat{k})\)

= \(4 \hat{i}-3 \hat{j}-12 \hat{k}\)

**Example 5. \(3 \hat{i}+4 \hat{j}+12 \hat{k}\) is a vector. Find the magnitude of the vector and the angles it makes with the x, y, and z axes.**

**Solution:**

If \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), then the magnitude \(|\vec{r}|=\sqrt{x^2+y^2+z^2}\)

∴ Magnitude of the vector \(3 \hat{i}+4 \hat{j}+12 \hat{k}\) is,

a = \(\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\)

Let the corresponding angles the vector makes with the x, y, and z axes be α, β, and γ respectively.

From definitions, \(\cos \alpha=\frac{x}{a}=\frac{3}{13} \text { or, } \alpha=\cos ^{-1} \frac{3}{13}\)

and \(\cos \beta=\frac{y}{a}=\frac{4}{13} \text { or, } \beta=\cos ^{-1} \frac{4}{13}\)

and \(\cos \gamma=\frac{z}{a}=\frac{12}{13} \text { or, } \gamma=\cos ^{-1} \frac{12}{13}\)

**Example 6. Two vectors \(\vec{A}\) and \(\vec{B}\) are \(\vec{A}=5 \hat{i}+3 \hat{j}-4 \hat{k}\) and \(\vec{B}=5 \hat{i}+2 \hat{j}+4 \hat{k}\). Find the unit vectors along \(\vec{A}+\vec{B}\) and \(\vec{A}-\vec{B}\).**

**Solution:**

⇒ \(\vec{A}=5 \hat{i}+3 \hat{j}-4 \hat{k} and \vec{B}=5 \hat{i}+2 \hat{j}+4 \hat{k}\)

∴ \(\vec{A}+\vec{B}=(5+5) \hat{i}+(3+2) \hat{j}+(-4+4) \hat{k}=10 \hat{i}+5 \hat{j}\)

and \(|\vec{A}+\vec{B}|=\sqrt{10^2+5^2}=5 \sqrt{5}\)

Hence, unit vector along \(\vec{A}+\vec{B}=\frac{\vec{A}+\vec{B}}{|\vec{A}+\vec{B}|}=\frac{2}{\sqrt{5}} \hat{i}+\frac{1}{\sqrt{5}} \hat{j}\)

Similarly, \(\vec{A}-\vec{B}=\hat{j}-8 \hat{k}\)

and \(|\vec{A}-\vec{B}|=\sqrt{1^2+(-8)^2}=\sqrt{65}\)

and unit vector along \(\vec{A}-\vec{B}=\frac{\vec{A}-\vec{B}}{|\vec{A}-\vec{B}|}=\frac{1}{\sqrt{65}} \hat{j}-\frac{8}{\sqrt{65}} \hat{k}\).

**Example 7. Two velocities \(\vec{v}_1\) and \(\vec{v}_2\) are 3 m/s towards north and 4 m/s towards east, respectively. Find \(\vec{v}_1-\vec{v}_2\).**

**Solution:**

Let us choose, the x-axis along the east and the y-axis along the north.

∴ \(\vec{v}_1=3 \hat{j} \mathrm{~m} / \mathrm{s}\) and \(\vec{v}_2=4 \hat{i} \mathrm{~m} / \mathrm{s}\)

∴ \(\vec{v}_1-\vec{v}_2=-4 \hat{i}+3 \hat{j}\) (between west and north)

∴ \(\left|\vec{v}_1-\vec{v}_2\right|=\sqrt{(-4)^2+3^2}=5 \mathrm{~m} / \mathrm{s}\)

If \(\vec{\nu}_1-\vec{v}_2\) makes an angle θ with east, then \(\tan \theta=\frac{3}{-4}=\tan 143.1^{\circ}=\tan \left(180^{\circ}-36.9^{\circ}\right)\)

∴ \(\vec{v}_1-\vec{v}_2\) is inclined at an angle of 143.1^{\circ} north of east, i.e., \(36.9^{\circ}\) north of west.

**Example 8. Find out the resultant of the following three displacement vectors: \(\vec{A}\) = 10 m, along north-west; \(\vec{B}\) = 20 m, 30° north of east; \(\vec{C}\) = 35 m, along the south.**

**Solution:**

Let us choose, x-axis along the east and the y-axis along the north. Taking into account the x- and y-components of the given vectors, they can be written as (in meter),

⇒ \(\vec{A}=\hat{i}\left(-10 \cos 45^{\circ}\right)+\hat{j} 10 \sin 45^{\circ}=-5 \sqrt{2} \hat{i}+5 \sqrt{2} \hat{j}\)

⇒ \(\vec{B}=\hat{i} 20 \cos 30^{\circ}+\hat{j} 20 \sin 30^{\circ}=10 \sqrt{3} \hat{i}+10 \hat{j}\)

⇒ \(\vec{C}=-35 \hat{j}\)

∴ Resultant, \(\vec{R}=\vec{A}+\vec{B}+\vec{C}\)

= \(\hat{i}(-5 \sqrt{2}+10 \sqrt{3})+\hat{j}(5 \sqrt{2}+10-35)\)

= \(\hat{i} 5(2 \sqrt{3}-\sqrt{2})-\hat{j} 5(5-\sqrt{2})\) [between east and south]

∴ R = \(|\vec{R}|=\sqrt{\{5(2 \sqrt{3}-\sqrt{2})\}^2+\{-5(5-\sqrt{2})\}^2}\)

= \(5 \sqrt{(2 \sqrt{3}-\sqrt{2})^2+(5-\sqrt{2})^2}=20.65 \mathrm{~m}\)

If \(\vec{R}\) makes an angle θ with the x-axis, then \(\tan \theta=\frac{R_y}{R_x}=\frac{-5(5-\sqrt{2})}{5(2 \sqrt{3}-\sqrt{2})}=-1.75=\tan \left(-60.2^{\circ}\right)\)

So, \(\vec{R}\) is inclined at \(60.2^{\circ}\) south of east.

**Example 9. The characteristic equations of a particle moving in a curved path are x = e ^{-t}, y = 2 cos 3t, and z = 2sin3t, where t stands for time. Find out**

**velocity and acceleration at any instant,****velocity and acceleration at t = 0.**

**Solution:**

Position vector of the particle at any instant, \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}=e^{-t} \hat{i}+2 \cos 3 \hat{t}+2 \sin 3 t \hat{k}\)

1. Velocity, \(\vec{v}=\frac{d \vec{r}}{d t}=-e^{-t} \hat{i}-6 \sin 3 t \hat{j}+6 \cos 3 t \hat{k}\)

= \(-x \hat{i}-3 z \hat{j}+3 y \hat{k}\)

Acceleration, \(\vec{a}=\frac{d \vec{v}}{d t}=e^{-t} \hat{i}-18 \cos 3 t \hat{j}-18 \sin 3 t \hat{k}\)

= \(x \hat{i}-9 y \hat{j}-9 z \hat{k}\)

2. At t = 0, \(\vec{v}=-e^0 \hat{i}-6 \sin (3 \times 0) \hat{j}+6 \cos (3 \times 0) \hat{k}\)

= \(-\hat{i}-6 \times 0 \hat{j}+6 \times 1 \hat{k}\)

= \(-\hat{i}+6 \hat{k}\)

v = \(|\vec{v}|=\sqrt{(-1)^2+6^2}=\sqrt{37}\)

At t = \(0, \vec{a}=e^0 \hat{i}-18 \cos (3 \times 0) \hat{j}-18 \sin (3 \times 0) \hat{k}\)

= \(\hat{i}-(18 \times 1) \hat{j}-(18 \times 0) \hat{k}\)

= \(\hat{i}-18 \hat{j}\)

a = \(|\vec{a}|=\sqrt{1^2+(-18)^2}=\sqrt{325}=5 \sqrt{13}\)

**Example 10. The position vector \(\vec{r}\) of a particle with respect to the origin changes with time t as \(\vec{r}=A t \hat{i}-B t^2 \hat{j}\), where A and B are positive constants. Determine**

**The locus of the particle,****The nature of variation with time of the velocity and acceleration vectors, and also the moduli of them.**

**Solution:**

1. \(\vec{r}=A t \hat{i}-B t^2 \hat{j}=x i+y \hat{j}\)

x = \(A t \quad \text { and } y=-B t^2\)

Then, \(x^2=A^2 t^2 or, t^2=\frac{x^2}{A^2}\)

also, \(t^2=-\frac{y}{B}\)

∴ \(\frac{x^2}{A^2}=-\frac{y}{B}\) or, \(x^2=-\frac{A^2}{B} y\)

This is the locus of the particle, which is a parabola.

2. Velocity, \(\vec{v}=\frac{d \vec{r}}{d t}=A \hat{i}-2 B t \hat{j}\)

Its modulus, \(|\vec{v}|=\sqrt{A^2+(-2 B t)^2}=\sqrt{A^2+4 B^2 t^2}\)

Acceleration, \(\vec{a}=\frac{d \vec{v}}{d t}=-2 B \hat{j}\)

Its modulus, \(|\vec{a}|=\sqrt{(-2 B)^2}=2 B\)

**Example 11. The position vector of a particle is, \(\vec{r}=3 t \hat{i}-2 t^2 \hat{j}+4 \hat{k}\). Find out**

**Its velocity \(\vec{v}\) and acceleration \(\vec{a}\),****The magnitude and direction of its velocity at t = 2 s.**

**Solution:**

∴ \(\vec{r}=3 t \hat{i}-2 t^2 \hat{j}+4 \hat{k}\)

1. \(\vec{v}=\frac{d \vec{r}}{d t}=3 \hat{i}-4 t \hat{j}\)

∴ \(\vec{a}=\frac{d \vec{v}}{d t}=-4 \hat{j}\)

2. At t = \(2 \mathrm{~s} \quad \vec{v}=3 \hat{i}-(4 \times 2) \hat{j}=3 \hat{i}-8 \hat{j}\) (between the x-axis and negative y-axis)

∴ \(|\vec{v}|=\sqrt{3^2+(-8)^2}=\sqrt{73} \text { unit }\)

Inclination of \(\vec{v}\) with the x-axis, \(\theta=\tan ^{-1}\left(\frac{-8}{3}\right)=-69.4^{\circ}\)