WBCHSE Class 12 Maths Solutions For Straight Line In Space

WBCHSE Class 12 Maths Solutions For Straight Line In Space Fundamental Concepts

This chapter consists of some important concepts of three-dimensional geometry. Though we have studied these in Class 9, yet we shall review here these fundamental concepts for ready reference.

Coordinates of a Point in Space

Let O be the origin, and let OX, OY, and OZ be three mutually perpendicular lines, taken as the x-axis, y-axis, and z-axis respectively in such a way that they form a right-handed system.

The planes YOZ, ZOX, and XOY are respectively known as the yz-plane, the zx-plane, and the xy-plane.

These planes, known as the coordinate planes, divide the space into eight parts called octants.

Let P be a point in space. Through P, draw three planes PLAN, PNBM, and PLCM parallel to the yz-plane, the zx-plane, and the xy-plane respectively, and meet the x-axis, y-axis, and z-axis at points A, B, and C respectively. Complete the parallelepiped whose coterminous edges are OA, OB, and OC.

Let OA = x, OB = y and OC = z. We say that the coordinates of P are (x,y,z).

It is clear from the figure alongside that

(1) x = distance of P from the yz-plane

(2) y = distance of P from the zx-plane

(3) z = distance of P from the xy-plane

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Also, we can say that

(1) the equation of the xy-plane is z = 0

(2) the equation of the zx-plane is y = 0

(3) the equation of the yz-plane is x = 0

Position Vector Of A Point In Space Let \(\hat{i}, \hat{j}, \hat{k}\) be unit vectors along OX, OY and OZ respectively.

If P(x, y, z) is a point in space, we say that the position vector (or, p.v.) of P is \((x \hat{i}+y \hat{j}+z \hat{k})\).

Some Results on Points in Space

1. Distance Between Two Points The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by PQ = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2} .\)

2. Sector Formulae

(1) If P(x,y,z) divides the join of A(x₁, y₁,z₁) and B(x₂,y₂,z₂) in the ration m:n then

\(x=\frac{\left(m x_2+n x_1\right)}{(m+n)}, y=\frac{\left(m y_2+n y_1\right)}{(m+n)}, z=\frac{\left(m z_2+n z_1\right)}{(m+n)} .\)

(2) The midpoint of the line joining A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

\(M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right) .\)

(3) The centroid of △ABC with vertices A(x₁,y₁,z₁), B(x₂,y₂,z₂) and C(x₃,y₃,z₃) is given by

\(G\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right) .\)

Some Results on Lines in Space

1. Direction Cosines Of A Line If a line makes angles α, β, γ with the x-axis, y-axis and z-axis respectively then

l = cos α, m = cos β, n = cos γ are called the direction cosines (or, d.c.’s)of the line.

We always have l² + m² + n² = 1.

Remarks

(1) d.c.’s of the x-axis are 1,0,0.

(2) d.c.’s of the y-axis are 0,1,0.

(3) d.c.’s of the z-axis are 0,0,1.

2. Direction Ratios Of A Line Any three numbers a, b, c proportional to the direction cosines l, m, n respectively of a line, are called the direction ratios of the line.

Clearly, we have \(\frac{l}{a}=\frac{m}{b}=\frac{n}{c} .\)

Some Important Facts

(1) If a,b,c are the direction ratios of a line then its direction cosines are

\(l=\frac{a}{\sqrt{a^2+b^2+c^2}}, m=\frac{b}{\sqrt{a^2+b^2+c^2}}, n=\frac{c}{\sqrt{a^2+b^2+c^2}} .\)

(2) If \(\vec{r}=a \hat{i}+b \hat{j}+c \hat{k}\) then the direction ratios of \(\vec{r}\) are a,b,c.

(3) Let PQ be a line joining P(x₁,y₁,z₁) and Q(x₂,y₂,z₂). Then the direction ratios of the line PQ are (x₂ – x₁), (y₂ – y₁), (z₂ – z₁).

3. Angle Between Two Lines If θ is the angle between two lines L₁ and L₂ whose d.c.’s are l₁, m₁, n₁ and l₂, m₂, n₂ then the following hold true.

(1) cos θ = l₁l₂ + m₁m₂ + n₁n₂

(2) sin θ = \(\sqrt{\Sigma\left(m_1 n_2-m_2 n_1\right)^2}\)

(3) lines L₁ and L₂ are prependicular ⇔ l₁l₂ + m₁m₂ + n₁n₂ = 0

(4) lines L₁ and L₂ are parallel ⇔ \(\frac{l_1}{l_2}=\frac{m_1}{m_2}=\frac{n_1}{n_2}\)

WBCHSE Class 12 Maths Solutions For Straight Line In Space Solved Examples

Example 1 Find the direction cosines of a line whose direction ratios are 2, -6, 3.

Solution

Here, a = 2, b = -6, c = 3.

∴ d.c.’s are \(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\),

i.e., \(\frac{2}{\sqrt{2^2+(-6)^2+3^2}}, \frac{-6}{\sqrt{2^2+(-6)^2+3^2}}, \frac{3}{\sqrt{2^2+(-6)^2+3^2}}\),

i.e., \(\frac{2}{7}, \frac{-6}{7}, \frac{3}{7} \text {. }\)

The direction cosines of a line \(\frac{2}{7}, \frac{-6}{7}, \frac{3}{7} \text {. }\)

Example 2 Find the direction cosines of each of the following vectors:

(1) \(2 \hat{i}+\hat{j}-2 \hat{k}\)

(2) \(-\hat{i}-\hat{k}\)

(3) \(-\hat{j}\)

Solution

(1) Direction ratios of the vector (\(2 \hat{i}+\hat{j}-2 \hat{k}\)) are 2, 1, -2.

∴ its d.c.’s are

\(\frac{2}{\sqrt{2^2+1^2+(-2)^2}}, \frac{1}{\sqrt{2^2+1^2+(-2)^2}}, \frac{-2}{\sqrt{2^2+1^2+(-2)^2}},\)

i.e., \(\frac{2}{3}, \frac{1}{3}, \frac{-2}{3} \text {. }\)

(2) Direction ratios of the vector (-\hat{i}-\hat{k}) are -1, 0, -1.

∴ its d.c.’s are

\(\frac{-1}{\sqrt{(-1)^2+0^2+(-1)^2}}, 0, \frac{-1}{\sqrt{(-1)^2+0^2+(-1)^2}}\)

i.e., \(\frac{-1}{\sqrt{2}}, 0, \frac{-1}{\sqrt{2}} \text {. }\)

(3) Direction ratios of the vector (-\hat{j}) are 0, -1, 0.

∴ its d.c.’s are

\(0, \frac{-1}{\sqrt{0^2+(-1)^2+0^2}},\) 0 i.e., 0, -1, 0.

Example 3 Find the direction cosines of the line segment joining the points A(7, -5, 9) and B(5, -3, 8).

Solution

Given

The points A(7, -5, 9) and B(5, -3, 8).

Direction ratios of the line segment AB are (5 – 7), -3 -(-5), (8 – 9), i.e., -2, 2, -1.

∴ its d.c.’s are

\(\frac{-2}{\sqrt{(-2)^2+2^2+(-1)^2}}, \frac{2}{\sqrt{(-2)^2+2^2+(-1)^2}}, \frac{-1}{\sqrt{(-2)^2+2^2+(-1)^2}}\)

i.e., \(\frac{-2}{3}, \frac{2}{3}, \frac{-1}{3} \text {. }\)

The direction cosines of the line segment joining the points \(\frac{-2}{3}, \frac{2}{3}, \frac{-1}{3} \text {. }\)

Example 4 Find the angle between the lines whose direction ratios are 2,3,6 and 1,2,2.

Solution

Given 2,3,6 and 1,2,2

Direction ratios of the first line are 2,3,6.

∴ its d.c.’s are

\(\frac{2}{\sqrt{2^2+3^2+6^2}}, \frac{3}{\sqrt{2^2+3^2+6^2}}, \frac{6}{\sqrt{2^2+3^2+6^2}}\)

i.e., \(\frac{2}{7}, \frac{3}{7}, \frac{6}{7} \text {. }\)

Direction ratios of the second line are 1, 2, 2.

∴ its d.c.’ are

\(\frac{1}{\sqrt{1^2+2^2+2^2}}, \frac{2}{\sqrt{1^2+2^2+2^2}}, \frac{2}{\sqrt{1^2+2^2+2^2}}\)

i.e., \(\frac{1}{3}, \frac{2}{3}, \frac{2}{3} \text {. }\)

∴ \(\cos \theta=\left(\frac{2}{7} \times \frac{1}{3}\right)+\left(\frac{3}{7} \times \frac{2}{3}\right)+\left(\frac{6}{7} \times \frac{2}{3}\right)=\frac{20}{21}\)

⇒ \(\theta=\cos ^{-1}\left(\frac{20}{21}\right)\)

Hence, the angle between the given lines is \(\cos ^{-1}\left(\frac{20}{21}\right)\).

Example 5 Find the angle between the vectors \(\overrightarrow{r_1}=3 \hat{i}-2 \hat{j}+\hat{k}\) and \(\overrightarrow{r_2}=4 \hat{i}+5 \hat{j}+7 \hat{k}\).

Solution

The direction ratios of the first vector are 3, -2, 1.

∴ its d.c.’s are

\(\frac{3}{\sqrt{3^2+(-2)^2+1^2}}, \frac{-2}{\sqrt{3^2+(-2)^2+1^2}}, \frac{1}{\sqrt{3^2+(-2)^2+1^2}}\)

i.e., \(\frac{3}{\sqrt{14}}, \frac{-2}{\sqrt{14}}, \frac{1}{14}\)

The direction ratios of the second vector are 4,5,7.

∴ its d.c.’s are

\(\frac{4}{\sqrt{4^2+5^2+7^2}}, \frac{5}{\sqrt{4^2+5^2+7^2}}, \frac{7}{\sqrt{4^2+5^2+7^2}}\)

i.e., \(\frac{4}{3 \sqrt{10}}, \frac{5}{3 \sqrt{10}}, \frac{7}{3 \sqrt{10}}\)

∴ cos θ = \(\left(\frac{3}{\sqrt{14}} \times \frac{4}{3 \sqrt{10}}\right)+\left(\frac{-2}{\sqrt{14}} \times \frac{5}{3 \sqrt{10}}\right)+\left(\frac{1}{\sqrt{14}} \times \frac{7}{3 \sqrt{10}}\right)\)

= \(\frac{9}{3 \sqrt{140}}=\frac{3}{\sqrt{140}}=\frac{3}{2 \sqrt{35}}\)

⇒ θ = \(\cos ^{-1}\left(\frac{3}{2 \sqrt{35}}\right)\)

Hence, the angle between the given vectors is \(\cos ^{-1}\left(\frac{3}{2 \sqrt{35}}\right)\).

Example 6 Find the angles made by the vector \(\vec{r}=(\hat{i}+\hat{j}-\hat{k})\) with the coordinate axes.

Solution

Direction ratios of the given vector are 1, 1, -1.

∴ its d.c.’s are

\(\frac{1}{\sqrt{1^2+1^2+(-1)^2}}, \frac{1}{\sqrt{1^2+1^2+(-1)^2}}, \frac{-1}{\sqrt{1^2+1^2+(-1)^2}}\)

i.e., \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\).

Let the given vector make angles θ₁, θ₂, and θ₃ with the x-axis, y-axis and z-axis respectively.

The d.c.’s of the x-axis are 1, 0, 0.

∴ \(\cos \theta_1=\left(\frac{1}{\sqrt{3}} \times 1\right)+\left(\frac{1}{\sqrt{3}} \times 0\right)+\left(-\frac{1}{\sqrt{3}} \times 0\right)=\frac{1}{\sqrt{3}}\)

⇒ \(\theta_1=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text {. }\)

∴ the angle between the given vector and the x-axis is \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\).

The d.c.’s of the y-axis are 0, 1, 0.

∴ \(\cos \theta_2=\left(\frac{1}{\sqrt{3}} \times 0\right)+\left(\frac{1}{\sqrt{3}} \times 1\right)+\left(-\frac{1}{\sqrt{3}} \times 0\right)=\frac{1}{\sqrt{3}}\)

⇒ \(\theta_2=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right).\)

∴ the angle between the given vector and the y-axis is \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)

The d.c.’s of the z-axis are 0,0,1.

∴ \(\cos \theta_3=\left(\frac{1}{\sqrt{3}} \times 0\right)+\left(\frac{1}{\sqrt{3}} \times 0\right)+\left(-\frac{1}{\sqrt{3}} \times 1\right)=\frac{-1}{\sqrt{3}}\)

⇒ \(\theta_3=\cos ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\)

∴ the angle between the given vector and the z-axis is \(\cos ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\)

WBCHSE Class 12 Maths Solutions For Straight Line In Space – Straight Line In Space

Equation of a Line Passing through a Given Point and Parallel to a Given Vector

Vector Form

Theorem 1 The vector equation of a straight line passing through a given point with position vector \(\overrightarrow{r_1}\) and parallel to a given vector \(\vec{m}\) is \(\vec{r}=\overrightarrow{r_1}+\lambda \vec{m}\), where λ is a scalar.

Proof

Let L be the line, passing through a given point A with position vector \(\overrightarrow{r_1}\) and parallel to a given vector \(\vec{m}\).

Let O be the origin. Then, \(\overrightarrow{O A}=\overrightarrow{r_1}\).

Let P be an arbitrary point on L, and let the position vector of P be \(\vec{r}\).

Then, \(\overrightarrow{O P}=\vec{r}\).

Clearly, \(\overrightarrow{A P} \ \vec{m}\)

⇒ \(\overrightarrow{A P}=\lambda \vec{m}\), for some scalar λ

⇒ (p.v. of P) – (p.v. of A) = \(\lambda \vec{m}\)

⇒ \(\overrightarrow{O P}-\overrightarrow{O A}=\lambda \vec{m}\)

⇒ \(\vec{r}-\overrightarrow{r_1}=\lambda \vec{m}\)

⇒ \(\vec{r}=\overrightarrow{r_1}+\lambda \vec{m}\) …(1)

Clearly, every point on the line L satisfies (1), and for any value of λ, (1) gives the position vector of a point P on the line.

Hence, \(\vec{r}=\vec{r}_1+\lambda \vec{m}\) is the desired equation.

Corollary The vector equation of a straight line passing through the origin and parallel to a given vector \(\vec{m}\) is \(\vec{r}=\lambda \vec{m}\).

Proof

Taking \(\overrightarrow{r_1}=\overrightarrow{0}\) in (1), we get the desired equation, \(\vec{r}=\lambda \vec{m}\).

Cartesian Form

Theorem 2 The equations of a straight line with direction ratios a, b, c, and passing through a point A(x₁, y_1, z₁) are

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} .\)

Proof

We know that the vector equation of a straight line passing through a fixed point A(x₁, y₁, z₁) with position vector \(\overrightarrow{r_1}\) and parallel to a given vector \(\vec{m}\) is given by

\(\vec{r}=\overrightarrow{r_1}+\lambda \vec{m}\) …(1), where λ is a scalar.

Let P(x, y, z) be the given point on the line with position vector \(\vec{r}\).

Then, \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k} \text { and } \overrightarrow{r_1}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text {. }\)

Since the direction ratios of the given line are a, b, c and this line is parallel to \vec{m} are a, b, c.

∴ \(\vec{m}=a \hat{i}+b \hat{j}+c \hat{k} .\)

Putting \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}, \vec{r}_1=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text { and } \vec{m}=a \hat{i}+b \hat{j}+c \hat{k}\) in (1), we get the equation of the line as

\((x \hat{i}+y \hat{j}+z \hat{k})=\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right)+\lambda(a \hat{i}+b \hat{j}+c \hat{k})\)

⇒ \((x \hat{i}+y \hat{j}+z \hat{k})=\left(x_1+\lambda a\right) \hat{i}+\left(y_1+\lambda b\right) \hat{j}+\left(z_1+\lambda c\right) \hat{k}\)

⇒ x = x₁ + λa, y = y₁ + λb and z = z₁ + λc

⇒ \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\lambda \text {. }\)

Hence, the equations of a line having direction ratios a, b, c and passing through A(x1, y1, z1) are

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}.\)

Corollary The equations of a line having direction cosines l, m, n and passing through (x1, y1, z1) are

\(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n} .\)

Proof Since the direction cosines of a line are porportional to the direction ratios of the line, the result follows.

Equation of a Line Passing through Two Given Points

Vector Form

Theorem 3 The vector equation of a straight line passing through two points with position vectors \(\overrightarrow{r_1} \text { and } \overrightarrow{r_2}\) is given by \(\vec{r}=\overrightarrow{r_1}+\lambda\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right) .\)

Proof

Let L be the given line, passing through two given points A and B with position vectors \(\overrightarrow{r_1} \text { and } \overrightarrow{r_2}\) respectively.

Let O be the origin.

Then, \(\overrightarrow{O A}=\overrightarrow{r_1} \text { and } \overrightarrow{O B}=\overrightarrow{r_2} \text {. }\)

∴ \(\overrightarrow{A B}=(\text { p.v. of } B)-(\text { p.v, of } A)=\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right) \text {. }\)

Let P be an arbitrary point on L, having the position vector \(\vec{r} .\)

Then, \(\overrightarrow{O P}=\vec{r}\).

∴ \(\overrightarrow{A P}=(\mathrm{p} \cdot \mathrm{v} \cdot \text { of } P)-(\mathrm{p} \cdot \mathrm{v} \cdot \text { of } A)\)

= \((\overrightarrow{O P}-\overrightarrow{O A})=\left(\vec{r}-\overrightarrow{r_1}\right)\)

Since \(\overrightarrow{A P} \text { and } \overrightarrow{A B}\) are collinear vectors, we have

\(\overrightarrow{A P}=\lambda(\overrightarrow{A B})\), for some scalar λ

⇒ \(\left(\vec{r}-\overrightarrow{r_1}\right)=\lambda\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right)\)

⇒ \(\vec{r}=\overrightarrow{r_1}+\lambda\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right)\)

Hence, the vector equation of a line L, passing through two given points A and B with position vectors \(\overrightarrow{r_1} \text { and } \vec{r}_2\), is given by

\(\vec{r}=\overrightarrow{r_1}+\lambda\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right) .\)

Cartesian Form

Theorem 4 The equation of a line passing through two given points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) are given by

\(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

Proof

We know that the vector equation of a line passing through two points A and B with position vectors \(\overrightarrow{r_1} \text { and } \overrightarrow{r_2}\) is given by

\(\vec{r}=\overrightarrow{r_1}+\lambda\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right)\) …(1)

Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) be the points on the given line with position vectors \(\overrightarrow{r_1} \text { and } \overrightarrow{r_2}\) respectively.

Let P(x, y, z) be an arbitrary point on this line with position vector \(\vec{r}\).

Then, \(\overrightarrow{r_1}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\),

\(\overrightarrow{r_2}=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k} \text { and } \vec{r}=x \hat{i}+y \hat{j}+z \hat{k} \text {. }\)

Substituting these values in (1), we get

\(x \hat{i}+y \hat{j}+z \hat{k}=\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right)+\lambda\left[\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\right]\)

⇒ \(\left(x-x_1\right) \hat{i}+\left(y-y_1\right) \hat{j}+\left(z-z_1\right) \hat{k}=\lambda\left[\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1 \hat{j}+\left(z_2-z_1\right) \hat{k}\right]\right.\)

⇒ \(\left(x-x_1\right)=\lambda\left(x_2-x_1\right),\left(y-y_1\right)=\lambda\left(y_2-y_1\right) \text { and }\left(z-z_1\right)=\lambda\left(z_2-z_1\right)\)

⇒ \(\frac{\left(x-x_1\right)}{\left(x_2-x_1\right)}=\frac{\left(y-y_1\right)}{\left(y_2-y_1\right)}=\frac{\left(z-z_1\right)}{\left(z_2-z_1\right)}(=\lambda)\), which are the required equatons.

Summary

1. (1) The vector equation of a line through a point with p.v. \(\overrightarrow{r_1}\) and parallel to \(\vec{m}\) is

\(\vec{r}=\overrightarrow{r_1}+\lambda \vec{m} .\)

(2) The vector equation of a line through two points with p.v.’s \(\overrightarrow{r_1} \text { and } \overrightarrow{r_2}\) is

\(\vec{r}=\overrightarrow{r_1}+\lambda\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right) .\)

2. (1) The Cartesian equations of a line with d.r.’s a, b, c and passing through A(x₁, y₁, z₁) are

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} .\)

(2) The Cartesian equations of a line through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) are

\(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

WBCHSE Class 12 Maths Solutions For Straight Line In Space Solved Examples

Example 1 A line passes through the point (3, 4, 5) and is parallel to the vector \((2 \hat{i}+2 \hat{j}-3 \hat{k})\). Find the equations of the line in the vector as well as Cartesian form.

Solution

Vector equation of the given line

The given line passes through the point A(3, 4, 5) and is parallel to the vector \(\vec{m}=(2 \hat{i}+2 \hat{j}-3 \hat{k})\).

Also, the p.v. of A is \(\overrightarrow{r_1}=3 \hat{i}+4 \hat{j}+5 \hat{k} .\)

∴ the vector equation of the given line is

\(\vec{r}=\overrightarrow{r_1}+\lambda \vec{m} \Leftrightarrow \vec{r}=(3 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+2 \hat{j}-3 \hat{k})\) …(1)

Cartesian equations of the given line

Taking \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), equation (1) becomes

\((x \hat{i}+y \hat{j}+z \hat{k})=(3 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+2 \hat{j}-3 \hat{k})\)

⇔ \((x \hat{i}+y \hat{j}+z \hat{k})=(3+2 \lambda) \hat{i}+(4+2 \lambda) \hat{j}+(5-3 \lambda) \hat{k}\)

⇔ x = 3 + 2λ, y = 4 + 2λ and z = 5 – 3λ

⇔ \(\frac{x-3}{2}=\frac{y-4}{2}=\frac{z-5}{-3}=\lambda .\)

Hence, \(\frac{x-3}{2}=\frac{y-4}{2}=\frac{z-5}{-3}\) are the required equations of the given line in the Cartesian form.

Example 2 A line is drawn in the direction of \((\hat{i}+\hat{j}-2 \hat{k})\) and it passes through a point with position vector \((2 \hat{i}-\hat{j}+4 \hat{k})\). Find the equations of the line in the vector as well as Cartesian form.

Solution

Vector equation of the given line

The given line passes through the point A having position vector \(\overrightarrow{r_1}=2 \hat{i}-\hat{j}+4 \hat{k}\), and is parallel to the vector \(\vec{m}=(\hat{i}+\hat{j}-2 \hat{k})\).

∴ the vector equation of the given line is

\(\vec{r}=\vec{r}_1+\lambda \vec{m}\)

⇔ \(\vec{r}=(2 \vec{i}-\vec{j}+4 \vec{k})+\lambda(\hat{i}+\hat{j}-2 \hat{k})\) …(1)

Cartesian equations of the given line

Taking \(\vec{r}=x \hat{i}+y \vec{j}+z \vec{k}\), equation (1) becomes

\((x \hat{i}+y \hat{j}+z \hat{k})=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(\hat{i}+\hat{j}-2 \hat{k})\)

⇔ \((x \hat{i}+y \hat{j}+z \hat{k})=(2+\lambda) \hat{i}+(\lambda-1) \hat{j}+(4-2 \lambda) \hat{k}\)

⇔ x = 2 + λ, y = λ – 1 and z = 4 – 2λ

⇔ \(\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-4}{-2}=\lambda\).

Hence, \(\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-4}{-2}\) are the required equations of the given line in the Cartesian form.

Example 3 Find the vector equation of a line passing through a point with the position vector \((2 \hat{i}-\hat{j}+\hat{k})\) and parallel to the line joining the points \((-\hat{i}+4 \hat{j}+\hat{k})\) and \((\hat{i}+2 \hat{j}+2 \hat{k})\). Also, find the Cartesian equivalents of the equation.

Solution

Vector equation of the given line

Let A, B, C be the given points with position vectors

\((2 \hat{i}-\hat{j}+\hat{k}),(-\hat{i}+4 \hat{j}+\hat{k}) \text { and }(\hat{i}+2 \hat{j}+2 \hat{k})\) respectively.

∴ \(\overrightarrow{B C}=(\text { p.v. of } C)-(\text { p.v. of } B)\)

= \((\hat{i}+2 \hat{j}+2 \hat{k})-(-\hat{i}+4 \hat{j}+\hat{k})=(2 \hat{i}-2 \hat{j}+\hat{k})\)

∴ the vector equation of the line passing through the point A with p.v. \((2 \hat{i}-\hat{j}+\hat{k})\), and parallel to \(\overrightarrow{B C}=(2 \hat{i}-2 \hat{j}+\hat{k})\), is

\(\vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\) …(1)

Cartesian equations of the given line

Taking \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), equation (1) becomes

\((x \hat{i}+y \hat{j}+z \hat{k})=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\)

⇔ \((x \hat{i}+y \hat{j}+z \hat{k})=(2+2 \lambda) \hat{i}+(-1-2 \lambda) \hat{j}+(1+\lambda) \hat{k}\)

⇔ x = 2 + 2λ, y = -1 -2λ and z = 1 + λ

⇔ \(\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-1}{1}=\lambda\)

Hence, \(\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-1}{1}\) are the required equations of the given line in the Cartesian form.

Example 4 Find the vector equation of the line passing through the point A(2, -1, 1), and parallel to the line joining the points B(-1, 4, 1) and C(1, 2, 2). Also, find the Cartesian equations of the line.

Solution

Vector equation of the given line

The p.v. of B = \((-\hat{i}+4 \hat{j}+\hat{k})\) and p.v. of C = \((\hat{i}+2 \hat{j}+2 \hat{k})\).

∴ \(\overrightarrow{B C}\) (p.v. of C) – (p.v. of B)

= \((\hat{i}+2 \hat{j}+2 \hat{k})-(-\hat{i}+4 \hat{j}+\hat{k})=(2 \hat{i}-2 \hat{j}+\hat{k})\)

The p.v. of A is \(\overrightarrow{r_1}=2 \hat{i}-\hat{j}+\hat{k} \text {. }\)

∴ the vector equation of the given line is

\(\vec{r}=\overrightarrow{r_1}+\lambda(\overrightarrow{B C})\)

⇔ \(\vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\) …(1)

Cartesian equations of the given line is

Taking \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), equation (1) becomes

\((x \hat{i}+y \hat{j}+z \hat{k})=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\)

⇔ \((x \hat{i}+y \hat{j}+z \hat{k})=(2+2 \lambda) \hat{i}+(-1-2 \lambda) \hat{j}+(1+\lambda) \hat{k}\)

⇔ x = 2 + 2λ, y = -1 -2λ, z = 1 + λ

⇔ \(\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-1}{1}=\lambda\)

Hence, \(\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-1}{1}\) are the required equations of the given line in the Cartesian form.

Example 5 The Cartesian equations of a line are \(\frac{x-3}{2}=\frac{y+2}{-5}=\frac{z-6}{3}\). Find the vector equation of the line.

Solution

The equations of the given line are

\(\frac{x-3}{2}=\frac{y+2}{-5}=\frac{z-6}{3}\).

This shows that the given line passes through the point A(3, -2, 6) and it is parallel to the vector \(\vec{m}=2 \hat{i}-5 \hat{j}+3 \hat{k}\).

Also, p.v. of A is \(\vec{r}_1=3 \hat{i}-2 \hat{j}+6 \hat{k} .\)

∴ the vector equation of the given line is

\(\vec{r}=\overrightarrow{r_1}+\lambda \vec{m}\)

⇔ \(\vec{r}=(3 \hat{i}-2 \hat{j}+6 \hat{k})+\lambda(2 \hat{i}-5 \hat{j}+3 \hat{k}) .\)

The vector equation of the line ⇔ \(\vec{r}=(3 \hat{i}-2 \hat{j}+6 \hat{k})+\lambda(2 \hat{i}-5 \hat{j}+3 \hat{k}) .\)

Example 6 The Cartesian equations of a line are 6x – 2 = 3y + 1 = 2z – 2.

Find (a) the direction ratios of the line, and (b) the Cartesian and vector equations of the line parallel to this line and passing through the point (2, -1, -1).

Solution

The equations of the given line are

6x – 2 = 3y + 1 = 2z – 2

⇔ \(\frac{\left(x-\frac{1}{3}\right)}{\left(\frac{1}{6}\right)}=\frac{\left(y+\frac{1}{3}\right)}{\left(\frac{1}{3}\right)}=\frac{z-1}{\left(\frac{1}{2}\right)}\)

⇔ \(\frac{\left(x-\frac{1}{3}\right)}{1}=\frac{\left(y+\frac{1}{3}\right)}{2}=\frac{z-1}{3}\).

(a) Clearly, the direction ratios of the given line are 1, 2, 3.

(b) The d.r.’s of a line parallel to the given line are 1, 2, 3.

∴ the Cartesian equations of the line passing through (2, -1, -1) and parallel to the given line are

\(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z+1}{3}\)

Vector equations of the required line

The required line passes throguh the point A(2, -1, -1) and is parallel to the vector \(\vec{m}=\hat{i}+2 \hat{j}+3 \hat{k} .\)

Also, p.v. of A is \(\vec{r}_1=2 \hat{i}-\hat{j}-\hat{k}\)

∴ the vector equation of the required line is

\(\vec{r}=\vec{r}_1+\lambda \vec{m}\),

i.e., \(\vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\).

The vector equation of the required line is \(\vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\).

Example 7 Find the vector and Cartesian equations of the line passing through the points A(2, -1, 4) and B(1, 1, -2).

Solution

Vector equation of the given line

Let the position vectors of A and B be \overrightarrow{r_1} \text { and } \overrightarrow{r_2} respectively. Then,

\(\overrightarrow{r_1}=2 \hat{i}-\hat{j}+4 \hat{k} \text { and } \vec{r}_2=\hat{i}+\hat{j}-2 \hat{k}\)

∴ \(\left(\vec{r}_2-\vec{r}_1\right)=(\hat{i}+\hat{j}-2 \hat{k})-(2 \hat{i}-\hat{j}+4 \hat{k})=(-\hat{i}+2 \hat{j}-6 \hat{k}) .\)

∴ the vector equation of the line AB is

\(\vec{r}=\vec{r}_1+\lambda\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right)\) for some scalar, λ,

i.e., \(\vec{r}=(2 \vec{i}-\vec{j}+4 \vec{k})+\lambda(-\vec{i}+2 \vec{j}-6 \vec{k})\) …(1)

Cartesian equations of the given line

Taking \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), equation (1) becomes

\((x \hat{i}+y \hat{j}+z \hat{k})=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(-\hat{i}+2 \hat{j}-6 \hat{k})\)

⇔ \((x \hat{i}+y \hat{j}+z \hat{k})=(2-\lambda) \hat{i}+(2 \lambda-1) \hat{j}+(4-6 \lambda) \hat{k}\)

⇔ x = 2 – λ, y 2λ – 1 and z = 4 – 6λ

⇔ \(\frac{x-2}{-1}=\frac{y+1}{2}=\frac{z-4}{-6}=\lambda\)

Hence \(\frac{x-2}{-1}=\frac{y+1}{2}=\frac{z-4}{-6}\) are the Cartesian equations of the given line.

NOTE A first-degree equation in the x, y, z of the form ax + by + cz + d = 0 represents a plane.

Example 8 Find the coordinates of the point where the line through A(3, 4, 1) and B(5, 1, 6) crosses the xy-plane.

Solution

A(3, 4, 1) and B(5, 1, 6)

The equations of the line AB are

\(\frac{x-3}{(5-3)}=\frac{y-4}{(1-4)}=\frac{z-1}{(6-1)} \text {, }\)

i.e., \(\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}\) …(1)

The line (1) crosses the xy-plane at the point where z = 0. So, putting z = 0 in (1), we get

\(x=\left(3-\frac{2}{5}\right)=\frac{13}{5}, y=\left(4+\frac{3}{5}\right)=\frac{23}{5} \text { and } z=0 \text {. }\)

Hence, the line AB crosses the xy-plane at \(\left(\frac{13}{5}, \frac{23}{5}, 0\right) \text {. }\)

Example 9 Find the coordinates of the point where the line \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}\) meets the plane 2x + 4y – z = 3.

Solution

The plane 2x + 4y – z = 3

Let \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}=\lambda\)

Then, x = (2λ + 1), y = (2 – 3λ), and z = (4λ – 5).

∴ P(2λ + 1, 2 – 3λ, 4λ – 5) is any point on (1).

If this point lies on the plane 2x + 4y – z = 3, we have

2(2λ+1) + 4(2-3λ) – (4λ-5) = 3 ⇒ λ = 1.

Putting λ = 1, we get the required point P(3, -1, -1).

Example 10 Find the equations of the line passing through the point (-1, 3, -2) and perpendicular to each of the lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5} \text {. }\)

Solution

(-1, 3, -2) and perpendicular to each of the lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x+2}{-3}=\frac{y-1}

Let the direction ratios of the required line be a, b, c. Then, this line being perpendicular to each of the given lines, we have

a + 2b + 3c = 0 …(1)

-3a + 2b + 5c = 0 …(2)

Cross multiplying (1) and (2), we get

[latex]\frac{a}{(10-6)}=\frac{b}{(-9-5)}=\frac{c}{(2+6)}\)

⇔ \(\frac{a}{4}=\frac{b}{-14}=\frac{c}{8}=k\) (say)

⇔ a = 4k, b = -14k and c = 8k.

Thus the required line has direction ratios 4k, -14k, 8k, and it passes through the point (-1, 3, -2).

Hence, the required equations are

\(\frac{x+1}{4 k}=\frac{y-3}{-14 k}=\frac{z+2}{8 k}\)

⇔ \(\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4} \text {. }\)

Example 11 Show that the lines \(\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5} \text { and } \frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}\) intersect each other. Also, find the point of their intersection.

Solution

The given lines are

\(\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}=\lambda\) (say) …(1)

and \(\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}=\mu\) (say) …(2)

P(4λ + 5, 4λ + 7, -5λ-3) is any point on (1).

Q(7μ + 8, μ + 3, 3μ + 5) is any point on (2).

The given lines will intersect if they have a common point. This happens when P and Q coincide for some particular values of λ and μ.

Thus, the given lines will intersect if

4λ + 5 = 7μ + 8, 4λ + 7 = μ + 3 and -5λ-3 = 3μ + 5

⇔ 4λ – 7 = 3 …(1), 4λ – μ = -3 …(2), 5λ + 3μ = -8 …(3)

Solving (1) and (2), we get λ = -1 and μ = -1.

Also, these values of λ and μ satisfy (3).

Hence, the given lines intersect.

Putting λ = -1 in P, or μ = -1 in Q, we get the point of intersection of the given lines as (, 3, ).

Example 12 Show that the lines \(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} \text { and } \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}\) do not intersect.

Solution

The given lines are

\(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda\) (say) …(1)

\(\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu\) (say) …(2)

P(3λ + 1, 2λ – 1, 5λ + 1)is any point on (1).

Q(4μ – , 3μ + 1, -2μ – 1) is any point on (2).

If possible, let the given lines intersect.

Then, P and Q coincide for some particular values of λ and μ.

In that case, we have

3λ + 1 = 4μ – 2, 2λ – 1 = 3μ + 1 and 5λ + 1 = -2μ – 1

⇔ 3λ – 4μ = -3 …(1), 2λ – 3μ = 2 …(2), 5λ + 2μ = -2 …(3).

Solving (1) and (2), we get λ = -17 and μ = -12.

However, these values of λ and μ do not satisfy (3).

Hence, the given lines do not intersect.

Example 13 Show that the lines \(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j}) \text { and } \vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+3 \hat{k})\) intersect each other. Find their point of intersection.

Solution

The given lines are

\(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j})\) …(1)

\(\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+3 \hat{k})\) …(2)

These lines will intersect if for some particular values of λ and μ, the values of \(\vec{r}\) given by (1) and (2) are the same,

i.e., \((\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j})=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+3 \hat{k})\) …(3)

Equating the coefficients of \(\hat{i}, \hat{j}, \hat{k}\) on both sides of (3), we get

1 + 3λ = 4μ + 2. 1 – λ = 0, and -1 = -1 + 3μ

⇔ 3λ – 2μ = 3 …(1), λ = 1 …(2), μ = 0 …(3).

Clearly, λ = 1 and μ = 0 also satisfy (1).

Putting λ = 1 in (1), we get \(\vec{r}=(4 \hat{i}+0 \hat{j}-\hat{k})\).

Hence, the point of intersection of the given lines is (4, 0, -1).

Example 14 Show that the lines \(\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \vec{r}=(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-\hat{j}+2 \hat{k})\) do not intersect.

Solution

The given lines are

\(\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) …(1)

\(\vec{r}=(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-\hat{j}+2 \hat{k})\) …(2)

These lines will intersect if for some particular values of λ and μ, the values of \(\vec{r}\) given by (1) and (2) are the same, i.e.,

\((\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})=(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-\hat{j}+2 \hat{k})\) …(3)

Equating the coefficients of \(\hat{i}, \hat{j}, \hat{k}\) on both sides of (3), we get

1 + λ = 1 + μ, 2 – λ = 1 – μ and 1 + λ = 1 + 2μ

⇒ λ – μ = 0 …(1), λ – μ = 1 …(2), λ – 2μ = 0 …(3).

From (2) and (3), we get λ = 2 and μ = 1.

And, these values of λ and μ do not satisfy (1).

Hence, the given lines do not intersect.

Example 15 Find the point on the line \(\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}\) at a distance 3√2 from the point (1, 2, 3).

Solution

The given line is

\(\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=r\) (say) …(1)

The general point on this line is

P(3r – 2, 2r – 1, 2r + 3).

Let this point P be at a distance 3√2 from the point Q(1, 2, 3).

Then, PQ = 3√2

⇒ PQ² = (3√2)² = 18

⇒ (3r – 2 – 1)² + (2r – 1 – 2)² + (2r + 3 – 3)² = 18

⇒ (3r – 3)² + (2r – 3)² + (2r)² = 18

⇒ 17r² + 18 – 30 r = 18

⇒ 17r² – 30r = 0 ⇒ r(17r – 30) = 0

⇒ r = 0 or r = \(\frac{30}{17}\).

r = 0 ⇒ required point is P(-2, -1, 3).

r = \(\frac{30}{17}\) ⇒ required point is P(\(\frac{90}{17}\)-2, \(\frac{60}{17}\)-1, \(\frac{60}{17}\)+3), i.e.,

P(\(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\)).

Example 16 Find the foot of the perpendicular drawn from the point P(1, 6, 3) on the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\). Also, find its distance from P.

Solution

The given line is

\(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=r\) (say) …(1)

The general point on this line is (r, 2r + 1, 3r + 2).

Let N be the foot of the perpendicular drawn from the point P(1, 6, 3) to the given line.

Then, this point is N(r, 2r + 1, 3r + 2) for some value of r.

DRs of PN are (r-1), (2r-5), (3r-1).

DRs of given line (1) are 1, 2, 3.

Since PN ⊥ given line (1), we have

1 . (r-1) + 2 . (2r-5) + 3 . (3r-1) = 0

⇔ (r + 4r + 9r) = 14 ⇔ 14r = 14 ⇔ r = 1.

So, the required point is N(1, 3, 5).

PN = \(\sqrt{(1-1)^2+(3-6)^2+(5-3)^2}\)

= \(\sqrt{0^2+(-3)^2+2^2}=\sqrt{13} \text { units. }\)

Hence, the required foot of the perpendicular is N(1, 3, 5) and its distance from P is √13 units.

Example 17 Find the image of the point (1, 6, 3) in the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} .\)

Solution

(1, 6, 3) in the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} .\)

The given line is

\(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda\) …(1)

Let N be the foot of the perpendicular drawn from the point P(1, 6, 3) to the given line.

∴ N has the coordinates (λ, 2λ + 1, 3λ + 2).

The direction ratios of PN are

(λ-1), (2λ + 1 – 6), (3λ + 2 – 3),

i.e., (λ-1), (2λ-5), (3λ-1).

Also, the direction ratios of the given line (1) are 1, 2, 3.

Since PN is perpendicular to the given line (1), we have

1 . (λ-1) + 2(2λ-5) + 3(3λ-1) = 0 ⇒ λ = 1.

Putting λ = 1, we get the point N(1, 3, 5).

Let M(α, β, γ) be the image of P(1, 6, 3) in the given line.

Then, N(1, 3, 5) is the midpoint of PM.

∴ \(\frac{\alpha+1}{2}=1, \frac{\beta+6}{2}=3 \text { and } \frac{\gamma+3}{2}=5\)

⇔ α = 1, β = 0 and γ = 7.

Hence, the image of P(1, 6, 3) is M(1, 0, 7).

Collinearity of Three Given Points

Theorem 1 The condition for three given points A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) to be collinear is that \(\frac{x_3-x_1}{x_2-x_1}=\frac{y_3-y_1}{y_2-y_1}=\frac{z_3-z_1}{z_2-z_1} .\)

Proof

A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃)

The equations of the line AB are

\(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\) …(1)

Clearly, A< B, C will be collinear only when C lies on the line Ab. This happens when C(x₃, y₃, z₃) lies on (1).

∴ \(\frac{x_3-x_1}{x_2-x_1}=\frac{y_3-y_1}{y_2-y_1}=\frac{z_3-z_1}{z_2-z_1} .\)

Collinearity of Three Points Whose p.v.’s are Given

Theorem 2 Three points A, B, C with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively are collinear if and only if there exist scalars μ₁, μ₂, μ₃, not all zero, such that \(\mu_1 \vec{a}+\mu_2 \vec{b}+\mu_3 \vec{c}=\overrightarrow{0} \text { and } \mu_1+\mu_2+\mu_3=0 \text {. }\)

Proof

Let A, B, C be three collinear points with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively.

Then, the vector equation of the line AB is

\(\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\)

⇒ \(\vec{c}=\vec{a}+\lambda(\vec{b}-\vec{a})\) [∵ A, B, C being collinear, C lies on AB]

⇒ \((1-\lambda) \vec{a}+\lambda \vec{b}-\vec{c}=\overrightarrow{0}\)

⇒ \(\mu_1 \vec{a}+\mu_2 \vec{b}+\mu_3 \vec{c}=\overrightarrow{0}\), where μ₁ + μ₂ + μ₃ = (1 – λ) + λ – 1 = 0 and μ₃ = -1 ≠ 0.

Conversely, let A, B, C be given points with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively and μ₁, μ₂, μ₃ be scalars, not all zero, say μ₃ ≠ 0, such that

\(\mu_1 \vec{a}+\mu_2 \vec{b}+\mu_3 \vec{c}=\overrightarrow{0} \text { and } \mu_1+\mu_2+\mu_3=0\)

⇒ \(\left(\frac{\mu_1}{\mu_3}\right) \vec{a}+\left(\frac{\mu_2}{\mu_3}\right) \vec{b}+\vec{c}=\overrightarrow{0} \text { and }\left(\frac{\mu_1}{\mu_3}\right)+\left(\frac{\mu_2}{\mu_3}\right)+1=0\)

⇒ \(\left(\frac{\mu_1}{\mu_3}\right) \vec{a}-\lambda \vec{b}+\vec{c}=\overrightarrow{0} \text { and }\left(\frac{\mu_1}{\mu_3}\right)-\lambda+1=0 \text {, where }\left(\frac{\mu_2}{\mu_3}\right)=-\lambda\)

⇒ \((\lambda-1) \vec{a}-\lambda \vec{b}+\vec{c}=\overrightarrow{0} [∵ \frac{\mu_1}{\mu_3}=(\lambda-1)]\)

⇒ \(\vec{c}=(1-\lambda) \vec{a}+\lambda \vec{b}\)

⇒ the point C lies on the line AB

⇒ the points A, B, C are collinear.

Solved Examples

Example 1 Prove that the points A(2, 1, 3), B(5, 0, 5) and C(-4, 3, -1) are collinear.

Solution

A(2, 1, 3), B(5, 0, 5) and C(-4, 3, -1)

The equations of the line AB are

\(\frac{x-2}{5-2}=\frac{y-1}{0-1}=\frac{z-3}{5-3}\)

⇔ \(\frac{x-2}{3}=\frac{y-1}{-1}=\frac{z-3}{2}\) …(1)

The given points A, B, C are collinear

⇔ C lies on the line AB

⇔ C(-4, 3, -1) satisfies (1)

⇔ \(\frac{-4-2}{3}=\frac{3-1}{-1}=\frac{-1-3}{2}\)

⇔ -2 = -2 = -2, which is true.

Hence, the given points A, B, C are collinear.

Example 2 Find the value of λ for which the points A(-1, 3, 2), B(-4, 2, -2) and C(5, 5, λ) are collinear.

Solution

A(-1, 3, 2), B(-4, 2, -2) and C(5, 5, λ)

The equations of the line AB are

\(\frac{x+1}{-4+1}=\frac{y-3}{2-3}=\frac{z-2}{-2-2}\)

⇔ \(\frac{x+1}{-3}=\frac{y-3}{-1}=\frac{z-2}{-4}\)

⇔ \(\frac{x+1}{3}=\frac{y-3}{1}=\frac{z-2}{4}\) …(1)

Since A, B, C are collinear, the points C(5, 5, λ) lies on (1).

∴ \(\frac{5+1}{3}=\frac{5-3}{1}=\frac{\lambda-2}{4} \Leftrightarrow \frac{\lambda-2}{4}=2 \Leftrightarrow \lambda=10 .\)

Hence, the value of λ is 10.

Example 3 Show that the points whose position vectors are \((-2 \hat{i}+3 \hat{j}+5 \hat{k}), (\hat{i}+2 \hat{j}+3 \hat{k}) \text { and }(7 \hat{i}-\hat{k})\) are collinear.

Solution

\((-2 \hat{i}+3 \hat{j}+5 \hat{k}), (\hat{i}+2 \hat{j}+3 \hat{k}) \text { and }(7 \hat{i}-\hat{k})\)

Let A, B, C be the given points, and let their position vectors be denoted by \(\vec{a}, \vec{b}, \vec{c}\) respectively. Then,

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

∴ the vector equation of the line AB is

\(\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\), where λ is a scalar

⇔ \(\vec{r}=(-2 \hat{i}+3 \hat{j}+5 \hat{k})+\lambda[(\hat{i}+2 \hat{j}+3 \hat{k})-(-2 \hat{i}+3 \hat{j}+5 \hat{k})]\)

⇔ \(\vec{r}=(-2 \hat{i}+3 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}-\hat{j}-2 \hat{k})\) …(1)

If this line passes through the point C, we have

\(\vec{c}=(-2 \hat{i}+3 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}-\hat{j}-2 \hat{k}) .\)

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

⇔ \((9 \hat{i}-3 \hat{j}-6 \hat{k})=\lambda(3 \hat{i}-\hat{j}-2 \hat{k})\)

⇔ \(3(3 \hat{i}-\hat{j}-2 \hat{k})=\lambda(3 \hat{i}-\hat{j}-2 \hat{k})\)

⇔ λ = 3, which is a scalar.

Thus, the point C lies on the line AB.

Hence, the given points A, B, C are collinear.

Example 4 Using the vector method, find the values of λ and μ for which the points A(3, λ, μ), B(2, 0, -3) and C(1, -2, -5) are collinear.

Solution

A(3, λ, μ), B(2, 0, -3) and C(1, -2, -5)

Let \(\vec{a}, \vec{b}, \vec{c}\) be the position vectors of the given points A, B, C respectively. Then,

\(\vec{a}=3 \hat{i}+\lambda \hat{j}+\mu \hat{k}, \vec{b}=2 \hat{i}-3 \hat{k} \text { and } \vec{c}=\hat{i}-2 \hat{j}-5 \hat{k}\)

∴ the vector equation of the line Bc is

\(\vec{r}=\vec{b}+t(\vec{c}-\vec{b})\), for some scalar t

⇔ \(\vec{r}=(2 \hat{i}-3 \hat{k})+t[(\hat{i}-2 \hat{j}-5 \hat{k})-(2 \hat{i}-3 \hat{k})]\)

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

⇔ \(\vec{r}=(2-t) \hat{i}-(2 t) \hat{j}-(3+2 t) \hat{k}\) …(1)

If this line BC passes through the point A, we have

\(\vec{a}=(2-t) \hat{i}-(2 t) \hat{j}-(3+2 t) \hat{k}\), for some scalar t

⇔ \((3 \hat{i}+\lambda \hat{j}+\mu \hat{k})=(2-t) \hat{i}-(2 t) \hat{j}-(3+2 t) \hat{k}\)

⇔ 2 – t = 3, -2t = λ and -(3 + 2t) = μ

⇔ t = -1, λ = (-2) x (-1) = 2, and μ = -[3 + 2 x(-1)] = -1.

Hence, λ = 2 and μ = -1.

Angle between Two Lines

Vector Form

Let the vector equations of two given lines be \(\vec{r}=\overrightarrow{r_1}+\lambda \overrightarrow{m_1} \text { and } \vec{r}=\overrightarrow{r_2}+\mu \overrightarrow{m_2}\), where λ and μ are scalars.

Let θ be the angle between these lines.

Since the given lines are parallel to \(vec{m}_1 \text { and } \vec{m}_2\) and respectively, the angle between the given lines must be equal to the angle between \(\vec{m}_1 \text { and } \vec{m}_2\).

∴ \(\cos \theta=\frac{\vec{m}_1 \cdot \vec{m}_2}{\left|\vec{m}_1\right| \cdot\left|\vec{m}_2\right|}\)

Cartesian Form

Let the Cartesian equations of two given lines be

\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \text { and } \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2} \text {. }\)

Then, the direction ratios of these lines are a₁, b₁, c₁ and a₂, b₂, c₂ respectively.

Let θ be the angle between these lines. Then,

\(\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)} .\)

Solved Examples

Example 1 Find the angle between the lines \(\vec{r}=(3 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(\hat{i}+2 \hat{j}+2 \hat{k}), and \vec{r}=(5 \hat{j}-2 \hat{k})+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k}) .\)

Solution

Let θ be the angle between the given lines. The given lines are parallel to the vectors \((\hat{i}+2 \hat{j}+2 \hat{k}) \text { and }(3 \hat{i}+2 \hat{j}+6 \hat{k})\) respectively.

So, θ is the angle between these vectors.

∴ \(\cos \theta=\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})}{|\hat{i}+2 \hat{j}+2 \hat{k}||3 \hat{i}+2 \hat{j}+6 \hat{k}|}\)

= \(\frac{(3+4+12)}{\left(\sqrt{1^2+2^2+2^2}\right)\left(\sqrt{3^2+2^2+6^2}\right)}=\frac{19}{21}\)

⇒ \(\theta=\cos ^{-1}\left(\frac{19}{21}\right)\)

Hence, the angle between the given lines is \(\cos ^{-1}\left(\frac{19}{21}\right)\).

Example 2 Find the angle between the lines \(\frac{x+4}{3}=\frac{y-1}{5}=\frac{z+3}{4} \text { and } \frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2} \text {. }\)

Solution

The direction ratios of the given lines are 3, 5, 4 and 1, 1, 2.

Let θ be the angle betweeen the given lines. Then,

\(\cos \theta=\frac{(3 \times 1+5 \times 1+4 \times 2)}{\left(\sqrt{3^2+5^2+4^2}\right)\left(\sqrt{1^2+1^2+2^2}\right)}=\frac{8}{5 \sqrt{3}}=\frac{8 \sqrt{3}}{15}\)

⇒ \(\theta=\cos ^{-1}\left(\frac{8 \sqrt{3}}{15}\right)\)

Hence, the angle between the given lines is \(\cos ^{-1}\left(\frac{8 \sqrt{3}}{15}\right)\)

Example 3 Find the angle between the lines \(\frac{x+1}{1}=\frac{2 y-3}{3}=\frac{z-6}{2} \text { and } \frac{x-4}{3}=\frac{y+3}{-2}, z=5 \text {. }\)

Solution

The given equations may be written as

\(\frac{x-4}{3}=\frac{y+3}{-2}=\frac{z-5}{0}\) …(1)

\(frac{x+1}{1}=\frac{y-(3 / 2)}{(3 / 2)}=\frac{z-6}{2}\) …(2)

Let \(\vec{m}_1 \text { and } \vec{m}_2\) be the vectors parallel to (1) and (2) respectively.

Then, \(\vec{m}_1=3 \hat{i}-2 \hat{j}+0 \hat{k} \text { and } \vec{m}_2=\hat{i}+\frac{3}{2} \hat{j}+2 \hat{k} \text {. }\)

Let θ be the angle between \(\vec{m}_1 \text { and } \vec{m}_2\), and so the angle between the given lines is also θ. Then,

\(\cos \theta=\frac{\vec{m}_1 \cdot \vec{m}_2}{\left|\overrightarrow{m_1}\right|\left|\vec{m}_2\right|}=\frac{(3 \times 1-2 \times 3 / 2+0 \times 2)}{\left(\sqrt{3^2+(-2)^2+0^2}\right)\left(\sqrt{1^2+(3 / 2)^2+2^2}\right)}=0\)

⇒ \(\theta=\frac{\pi}{2}\)

Hence, the angle between the given lines is \(\left(\frac{\pi}{2}\right)\).

Example 4 Find the angle between the lines \(\frac{5-x}{3}=\frac{y+3}{-4}, z=7 \text { and } \frac{x}{1}=\frac{1-y}{2}=\frac{z-6}{2} \text {. }\)

Solution

The given lines are

\(\frac{x-5}{-3}=\frac{y+3}{-4}=\frac{z-7}{0}\) …(1)

\(\frac{x}{1}=\frac{y-1}{-2}=\frac{z-6}{2}\) …(2)

Let \(\vec{m}_1 \text { and } \vec{m}_2\) be the vectors parallel to (1) and (2) respectively.

Then, \(\vec{m}_1=-3 \hat{i}-4 \hat{j}+0 \hat{k} \text { and } \vec{m}_2=\hat{i}-2 \hat{j}+2 \hat{k}\)

Let θ be the angle between \(\vec{m}_1 \text { and } \vec{m}_2\). Then,

\(\cos \theta=\frac{\vec{m}_1 \cdot \vec{m}_2}{\left|\vec{m}_1\right|\left|\vec{m}_2\right|}\)

= \(\frac{\{(-3) \times 1+(-4) \times(-2)+0 \times 2\}}{\left\{\sqrt{(-3)^2+(-4)^2+0^2}\right\}\left\{\sqrt{1^2+(-2)^2+2^2}\right\}}=\frac{1}{3}\)

⇒ \(\theta=\cos ^{-1}\left(\frac{1}{3}\right)\)

Hence, the angle between the given lines is \(\cos ^{-1}\left(\frac{1}{3}\right)\).

Example 5 Find the angle between two lines, one of which has direction ratios 2, 2, 1, and the other is obtained by joining the points (3, 1, 4) and (7, 2, 12).

Solution

Let L₁ and L₂ be the given lines. Then, the direction ratios of L₁ are 2, 2, 1.

And, the direction ratios of L₂ are (7 – 3), (2 – 1), (12 – 4), i.e., 4, 1, 8.

Let θ be the angle between the given lines. Then,

\(\cos \theta=\frac{(2 \times 4+2 \times 1+1 \times 8)}{\left(\sqrt{2^2+2^2+1^2}\right)\left(\sqrt{4^2+1^2+8^2}\right)}=\frac{18}{27}=\frac{2}{3}\)

⇒ \(\theta=\cos ^{-1}\left(\frac{2}{3}\right)\)

Hence, the angle between the given lines is \(\cos ^{-1}\left(\frac{2}{3}\right)\).

Example 6 Prove that the lines x = ay + b, z = cy + d, and x = a’y + b’, z = c’y + d’ are perpendicular if aa’ + cc’ + 1 = 0.

Solution

The equations of the first line are

x = ay + b, z = cy + d

⇔ \(\frac{x-b}{a}=y, \frac{z-d}{c}=y\)

⇔ \(\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}\) …(1)

Similarly, the equations of the second line are

\(\frac{x-b^{\prime}}{a^{\prime}}=\frac{y}{1}=\frac{z-d^{\prime}}{c^{\prime}}\) …(2)

The given lines are perpendicular to each other

⇔ aa’ + 1 x + cc’ = 0

⇔ aa’ + cc’ + 1 = 0.

Example 7 Find the value of k so that the lines \(\frac{1-x}{3}=\frac{7 y-14}{2 k}=\frac{z-3}{2} \text { and } \frac{7-7 x}{3 k}=\frac{5-y}{1}=\frac{6-z}{5}\) are at right angles.

Solution

The given equations in the standard form are

\(\frac{x-1}{-3}=\frac{y-2}{(2 k / 7)}=\frac{z-3}{2}\) …(1)

\(\frac{x-1}{(-3 k / 7)}=\frac{y-5}{-1}=\frac{z-6}{-5}\) …(2)

The direction ratios of the given lines are

\(\left(a_1=-3, b_1=\frac{2 k}{7}, c_1=2\right) \text { and }\left(a_2=\frac{-3 k}{7}, b_2=-1, c_2=-5\right)\)

The given lines are at right angles

⇔ a₁a₂ + b₁b₂ + c₁c₂ = 0

⇔ \(-3 \times\left(\frac{-3 k}{7}\right)+\left(\frac{2 k}{7}\right) \times(-1)+2 \times(-5)=0\)

⇔ \(\frac{9 k}{7}-\frac{2 k}{7}=10 \Leftrightarrow 7 k=70 \Leftrightarrow k=10 .\)

Hence, k = 10.

Miscellaneous Solved Examples

Example 1 The direction ratios of a vector are 2, -3, 4. Find its direction cosines.

Solution

2, -3, 4

Clearly, the direction cosines of the given vector are

\(\frac{2}{\sqrt{2^2+(-3)^2+4^2}}, \frac{-3}{\sqrt{2^2+(-3)^2+4^2}}, \frac{4}{\sqrt{2^2+(-3)^2+4^2}}\),

i.e., \(\frac{2}{\sqrt{29}}, \frac{-3}{\sqrt{29}}, \frac{4}{\sqrt{29}} .\)

Example 2 Find the direction cosines of the vector \(\vec{r}=6 \hat{i}+2 \hat{j}-3 \hat{k}\).

Solution

The direction ratios of \vec{r} are 6, 2, -3.

∴ the direction cosines of \(\vec{r}\) are

\(\frac{6}{\sqrt{6^2+2^2+(-3)^2}}, \frac{2}{\sqrt{6^2+2^2+(-3)^2}}, \frac{-3}{\sqrt{6^2+2^2+(-3)^2}}\)

i.e., \(\frac{6}{7}, \frac{2}{7}, \frac{-3}{7} \text {. }\)

Example 3 If a line makes angles α, β, γ with the x-axis, y-axis and z-axis respectively, prove that sin²α + sin²β + sin²γ = 2.

Solution

The direction cosines of the line are cos α, cos β, cos γ.

∴ cos²α + cos²β + cos²γ = 1

⇒ (1 – sin²α) + (1 – sin²β) + (1 – sin²γ) = 1

⇒ sin²α + sin²β + sin²γ = 2.

Example 4 Find the angle between the vectors having direction cosines \(\frac{1}{2},-\frac{1}{3}, \frac{1}{4}, \text { and } \frac{1}{3}, 1, \frac{2}{3} \text {. }\)

Solution

Let θ be the angle between the given vectors. Then,

\(\cos \theta=\frac{\left\{\frac{1}{2} \times \frac{1}{3}+\left(-\frac{1}{3}\right) \times 1+\frac{1}{4} \times \frac{2}{3}\right\}}{\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{-1}{3}\right)^2+\left(\frac{1}{4}\right)^2} \cdot \sqrt{\left(\frac{1}{3}\right)^2+(1)^2+\left(\frac{2}{3}\right)^2}}=0\)

⇒ θ = 90°.

Example 5 Find the angle between the vectors having direction ratios 3, 4, 5, and 4, -3, 5.

Solution

3, 4, 5, and 4, -3, 5

Let θ be the angle between the given vectors.

Let a₁ = 3, b₁ = 4, c₁ = 5, and a₂ = 4, b₂ = -3, c₂ = 5.

\(\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)}\)

= \(\frac{3 \times 4+4 \times(-3)+5 \times 5}{\left(\sqrt{3^2+4^2+5^2}\right)\left(\sqrt{4^2+(-3)^2+5^2}\right)}=\frac{25}{50}=\frac{1}{2}\)

∴ θ = 60°.

Thus, the angle between the given vectors is 60°.

Example 6 Find the angles of △ABC whose vertices are A(-1, 3, 2), B(2, 3, 5) and C(3, 5, -2).

Solution

A(-1, 3, 2), B(2, 3, 5) and C(3, 5, -2)

The d.r.’s of AB are (2+1), (3-3), (5-2), i.e., 3, 0, 3.

And, the d.r.’s of AC are (3+1), (5-3), (-2-2), i.e., 4, 2, -4.

∴ \(\cos A=\frac{[3 \times 4+0 \times 2+3 \times(-4)]}{\left(\sqrt{3^2+0^2+3^2}\right)\left(\sqrt{4^2+2^2+(-4)^2}\right)}=0 \Rightarrow \angle A=90^{\circ} \text {. }\)

The d.r.’s of BA are (-1-2), (3-3), (2-5), i.e., -3, 0, -3.

And, the d.r.’s of BC are (3-2), (5-3), (-2-5), i.e., 1, 2, -7.

∴ \(\cos B=\frac{(-3) \times 1+0 \times 2+(-3) \times(-7)}{\left(\sqrt{(-3)^2+0^2+(-3)^2}\right)\left(\sqrt{1^2+2^2+(-7)^2}\right)}=\frac{1}{\sqrt{3}}\)

⇒ \(\angle B=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right).\)

The d.r.’s of CB are (2-3),(3-5), (5+2) i.e., -1, -2, 7.

And, the d.r.’s of CA are (-1-3), (3-5), (2+2), i.e., -4, -2, 4.

∴ \(\cos C=\frac{(-1) \times(-4)+(-2) \times(-2)+7 \times 4}{\left\{\sqrt{(-1)^2+(-2)^2+7^2}\right\}\left\{\sqrt{(-4)^2+(-2)^2+4^2}\right\}}=\sqrt{\frac{2}{3}} .\)

⇒ \(C=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right) .\)

Thus, ∠A = 90°, ∠B = \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text { and } \angle C=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right) \text {. }\)

Example 7 Find the direction cosines of the line which is perpendicular to thr lines whose direction ratios are 1, -1, 2 and 2, 1, -1.

Solution

1, -1, 2 and 2, 1, -1

Let l, m, n be the direction cosines of the required line.

Then, l x 1 + m x (-1) + n x 2 = 0,

l x 2 + m x 1 + n x (-1) = 0

⇒ \(\left\{\begin{array}{c}
l-m+2 n=0 \\
2 l+m-n=0
\end{array}\right.\)

Cross multiplying those lines, we get

\(\frac{l}{-1}=\frac{m}{5}=\frac{n}{3}=\frac{\sqrt{l^2+m^2+n^2}}{\sqrt{(-1)^2+5^2+3^2}}=\frac{1}{\sqrt{35}}\)

∴ \(l=\frac{-1}{\sqrt{35}}, m=\frac{5}{\sqrt{35}}, n=\frac{3}{\sqrt{35}} .\)

Example 8 Find the direction cosines of the lines which are connected by the relations l – 5m + 3n = 0 and 7l² + 5m2 – 3n² = 0.

Solution

The given equations are

l – 5m + 3n = 0 …(1)

7l² + 5m² – 3n² = 0 …(2)

Putting l = (5m – 3n) from (1) in (2), we get

7(5m – 3n)² + 5m² – 3n² = 0

⇒ 6m² – 7mn + 2n² = 0

⇒ \(6\left(\frac{m}{n}\right)^2-7\left(\frac{m}{n}\right)+2=0 \Rightarrow 6 p^2-7 p+2=0, \text { where } \frac{m}{n}=p\)

⇒ (3p – 2)(2p – 1) = 0

⇒ \(p=\frac{2}{3} \text { or } p=\frac{1}{2} \Rightarrow \frac{m}{n}=\frac{2}{3} \text { or } \frac{m}{n}=\frac{1}{2} \text {. }\)

Now, \(\frac{m}{n}=\frac{2}{3} \Rightarrow \frac{m}{2}=\frac{n}{3}=\frac{5 m-3 n}{5 \times 2-3 \times 3}=\frac{l}{1}\)

⇒ \(\frac{l}{1}=\frac{m}{2}=\frac{n}{3}=\frac{\sqrt{l^2+m^2+n^2}}{\sqrt{1^2+2^2+3^2}}=\frac{1}{\sqrt{14}}\)

⇒ \(l=\frac{1}{\sqrt{14}}, m=\frac{2}{\sqrt{14}}, n=\frac{3}{\sqrt{14}} .\)

Again, \(\frac{m}{n}=\frac{1}{2} \Rightarrow \frac{m}{1}=\frac{n}{2}=\frac{5 m-3 n}{5 \times 1-3 \times 2}=\frac{l}{-1}\)

⇒ \(\frac{l}{-1}=\frac{m}{1}=\frac{n}{2}=\frac{\sqrt{l^2+m^2+n^2}}{\sqrt{(-1)^2+1^2+2^2}}=\frac{1}{\sqrt{6}}\)

⇒ \(l=\frac{-1}{\sqrt{6}}, m=\frac{1}{\sqrt{6}}, n=\frac{2}{\sqrt{6}} .\)

Hence, the direction cosines of the lines are

\(\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right) \text { and }\left(\frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right) \text {. }\)

Example 9 Prove that the straight lines whose direction cosines are given by the relations al + bm + cn = 0 and fmn + gnl + hlm = 0 are perpendicular to each other if \(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0\), and parallel if a²f² + b²g² + c²h² – 2bcgh – 2cahf – 2abfg = 0.

Solution

The given equations are

al + bm + cn = 0 …(1)

fmn + gnl + hlm = 0 …(2)

Putting \(n=\frac{-(a l+b m)}{c}\) from (1) in (2), we get

\(f m \cdot\left\{\frac{-(a l+b m)}{c}\right\}+g l \cdot\left\{\frac{-(a l+b m)}{c}\right\}+h l m=0\)

⇒ agl² + (af + bg – ch)lm + bfm² = 0

⇒ \(a g\left(\frac{l}{m}\right)^2+(a f+b g-c h) \cdot\left(\frac{l}{m}\right)+b f=0\) …(3)

Now, equation (3), being a quadratic equation in \(\left(\frac{l}{m}\right)\), will have two roots, say \(\left(\frac{l_1}{m_1}\right) \text { and }\left(\frac{l_2}{m_2}\right)\).

∴ \(\frac{l_1}{m_1} \times \frac{l_2}{m_2}=\frac{b f}{a g} \Rightarrow \frac{l_1 l_2}{b f}=\frac{m_1 m_2}{a g}\)

⇒ \(\frac{l_1 l_2}{\left(\frac{f}{a}\right)}=\frac{m_1 m_2}{\left(\frac{g}{b}\right)}=\frac{n_1 n_2}{\left(\frac{h}{c}\right)}=k\) [by symmetry].

∴ \(l_1 l_2+m_1 m_2+n_1 n_2=k\left(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\right)\)

Thus, the given lines will be perpendicular to each other

⇔ l₁l₂ + m₁m₂ + n₁n₂ = 0

⇔ \(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0 .\)

The lines will be parallel only when the roots of (3) are equal.

∴ (af + bg – ch)² – 4abgf = 0

⇔ a²f² + b²g² + c²h² – 2bcgh – 2cahf – 2abfg = 0.

Example 10 Show that the straight lines whose direction cosines are given by the equations al + bm + cn = 0 and ul² + vm² + wn² = 0 are mutually perpendicular if a²(v + w) + b²(u+w) + c²(u+v) = 0, and parallel if \(\frac{a^2}{u}+\frac{b^2}{v}+\frac{c^2}{w}=0 .\)

Solution

The given equations are

al + bm + cn = 0 …(1)

ul² + vm² + wn² = 0 …(2)

Putting \(l=\frac{-(b m+c n)}{a}\) from (1) in (2), we get

\(\frac{u(b m+c n)^2}{a^2}+v m^2+w n^2=0\)

⇒ (b²u + a²v)m² + 2ubcmn + (c²u + a²w)n² = 0

⇒ \(\left(b^2 u+a^2 v\right)\left(\frac{m}{n}\right)^2+2 u b c\left(\frac{m}{n}\right)+\left(c^2 u+a^2 w\right)=0\) …(3)

Let \(\frac{m_1}{n_1} \text { and } \frac{m_2}{n_2}\) be the roots of (3).

Then, \(\frac{m_1}{n_1}, \frac{m_2}{n_2}=\frac{c^2 u+a^2 w}{b^2 u+a^2 v}\)

⇒ \(\frac{m_1 m_2}{c^2 u+a^2 w}=\frac{n_1 n_2}{b^2 u+a^2 v}=\frac{l_1 l_2}{b^2 w+c^2 v}=k\) [by symmetry]

∴ l₁l₂ + m₁m₂ + n₁n₂ = k(b²w + c²v + c²u + a²w + b²u + a²v)

The given lines are mutually perpendicular

⇔ l₁l₂ + m₁m₂ + n₁n₂ = 0

⇔ a²(v+w) + b²(w+u) + c²(u+v)= 0.

For the given lines to be parallel, the direction cosines must be equal.

∴ the roots of (3) must be equal.

∴ \(4 u^2 b^2 c^2-4\left(b^2 u+a^2 v\right)\left(c^2 u+a^2 w\right)=0 \Leftrightarrow \frac{a^2}{u}+\frac{b^2}{v}+\frac{c^2}{w}=0 .\)

Example 11 If the edges of a rectangular parallelepiped are a, b, c, prove that the angles between the four diagonals are given by \(\cos ^{-1}\left(\frac{ \pm a^2 \pm b^2 \pm c^2}{a^2+b^2+c^2}\right)\).

Solution

Let OA, OB, OC be the coterminous edges of the parallelepiped, taken along the axes in such a way that OA = a, OB = b and OC = c.

Then, the coordinates of the vertices are O(0,0,0), A(a,0,0), B(0,b,0), C(0,0,c), P(a,b,c), L(0,b,c), M(a,0,c) and N(a,b,0).

The direction ratios of the diagonals OP, AL, BM and CN are (a,b,c), (-a,b,c), (a,-b,c) and (a,b,-c) respectively.

∴ the direction cosines of OP, AL, BM and CN are

\(\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)\),

\(\left(\frac{-a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)\),

\(\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{-b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)\),

\(\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{-c}{\sqrt{a^2+b^2+c^2}}\right)\).

Let θ₁ be the angle between OP and AL. Then,

\(\cos \theta_1=\frac{\left(-a^2+b^2+c^2\right)}{\left(a^2+b^2+c^2\right)} \quad \text { or } \theta_1=\cos ^{-1}\left(\frac{-a^2+b^2+c^2}{a^2+b^2+c^2}\right)\)

Again, let θ₂ be the angle between OP and BM. Then,

\(\cos \theta_2=\frac{\left(a^2-b^2+c^2\right)}{\left(a^2+b^2+c^2\right)} \quad \text { or } \theta_2=\cos ^{-1}\left(\frac{a^2-b^2+c^2}{a^2+b^2+c^2}\right) \text {. }\)

Similarly, the angles between the other pairs of diagonals can be obtained.

Clearly, the angles between the four diagonals can be given by

\(\cos ^{-1}\left(\frac{ \pm a^2 \pm b^2 \pm c^2}{a^2+b^2+c^2}\right) \text {. }\)

Example 12 Show that the angle between any two diagonals of a cube is \(\cos ^{-1}\left(\frac{1}{3}\right)\).

Solution

Let OA, OB, OC be the coterminous edges of a cube, taken along the axes in such a way that OA = OB = OC = a.

Then, the coordinates of the vertices of the cube are

O(0,0,0), A(a,0,0), B(0,a,0), C(0,0,a), P(a,a,a), L(0,a,a), M(a,0,a) and N(a,a,0).

The direction ratios of the diagonals OP, AL, BM and CN are (a,a,a), (-a,a,a), (a,-a,a) and (a,a,-a) respectively.

Thus, direction cosines of OP, AL, BM and CN are

\(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right), and \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)\) respectively.

If θ₁ be the angle between OP and AL then

\(\cos \theta_1=\left\{\frac{1}{\sqrt{3}} \cdot\left(\frac{-1}{\sqrt{3}}\right)+\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}\right\}=\frac{1}{3}\)

⇒ \(\theta_1=\cos ^{-1}\left(\frac{1}{3}\right)\)

Similarly, the angle between each one of the other pairs is \(\cos ^{-1}\left(\frac{1}{3}\right)\).

Hence, the angle between any two diagonals of the cube is \(\cos ^{-1}\left(\frac{1}{3}\right)\).

Example 13 A line makes angles α, β, γ, δ with the four diagonals of a cube. Prove that \(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta=\frac{4}{3} \text {. }\).

Solution

Let OA, OB, OC be the coterminous edges of a cube, taken along the axes in such a way that OA = OB = OC = a.

Then, the coordinates of the vertices of the cube are O(0,0,0), A(a,0,0), B(0,a,0), C(0,0,a), P(a,a,a), L(0,a,a), M(a,0,a) and N(a,a,0).

The direction ratios of the diagonals OP, AL, BM and CN are (a,a,a), (-a,a,a), (a,-a,a) and (a,a,-a) respectively.

∴ the direction cosines of OP, AL, BM and CN are

\(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\), and \(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)\) respectively.

Let (l,m,n) be the direction cosines of a line which makes angles α, β, γ, δ with the four diagonals of the cube. Then,

\(\cos \alpha=\left(l \cdot \frac{1}{\sqrt{3}}+m \cdot \frac{1}{\sqrt{3}}+n \cdot \frac{1}{\sqrt{3}}\right)=\frac{(l+m+n)}{\sqrt{3}}\),

\(\cos \beta=\left\{l \cdot\left(\frac{-1}{\sqrt{3}}\right)+m \cdot \frac{1}{\sqrt{3}}+n \cdot \frac{1}{\sqrt{3}}\right\}=\frac{(-l+m+n)}{\sqrt{3}}\),

\(\cos \gamma=\left\{l \cdot \frac{1}{\sqrt{3}}+m \cdot\left(\frac{-1}{\sqrt{3}}\right)+n \cdot \frac{1}{\sqrt{3}}\right\}=\frac{(l-m+n)}{\sqrt{3}}\),

\(\cos \delta=\left\{l \cdot \frac{1}{\sqrt{3}}+m \cdot \frac{1}{\sqrt{3}}+n \cdot\left(\frac{-1}{\sqrt{3}}\right)\right\}=\frac{(l+m-n)}{\sqrt{3}}\).

On squaring and adding, we get

\(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta\)

= \(\frac{1}{3} \cdot\left\{(l+m+n)^2+(-l+m+n)^2+(l-m+n)^2+(l+m-n)^2\right\}\)

= \(\frac{4}{3}\).

Example 14 If a variable line in two adjacent positions has direction cosines (l, m, n) and (l + δl, ,m + δm, n + δn), show that the small angle δθ between the two positions is given by (δθ)² = (δl)² + (δm)² + (δn)².

Solution

Clearly, we have

l² + m² + n² = 1 …(1)

and (l + δl)² + (m + δm)² + (n + δn)² = 1 …(2)

Subtracting (1) from (2), we get

(l + δl)² + (m + δm)² + (n + δn)² – (l² + m² + n²) = 0

⇒ (δl)² + (δm)² + (δn)² = -2(l . δl + m .δm + n . δn) …(3)

∴ cos δθ = l . (l + δl) + m . (m + δm) + n . (n + δn)

= (l² + m² + n²) + (l . δl + m . δm + n . δn)

= \(1-\frac{1}{2}\left\{(\delta l)^2+(\delta m)^2+(\delta n)^2\right\}\) [using (3)].

∴ (δl)² + (δm)² + (δn)² = 2(1 – cos δθ)

= \(4 \sin ^2 \frac{\delta \theta}{2}=4 \cdot\left(\frac{\delta \theta}{2}\right)^2 \quad\left[\frac{\delta \theta}{2} \text { being small, } \sin \frac{\delta \theta}{2}=\frac{\delta \theta}{2}\right]\)

= (δθ)².

Hence, (δθ)² = (δl)² + (δm)² + (δn)².

Example 15 If l₁, m₁, n₁, and l₂, m₂, n₂ be the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of them are (m₁n₂ – m₂n₁), (n₁l₂ – n₂l₁), (l₁m₂ – l₂m₁).

Solution

Let l, m, n be the direction cosines of the line perpendicular to each of the given lines. Then,

ll₁ + mm₁ + nn₁ = 0 …(1)

ll₂ + mm₂ + nn₂ = 0 …(2)

Cross multiplying (1) and (2), we get

\(\frac{l}{\left(m_1 n_2-m_2 n_1\right)}=\frac{m}{\left(n_1 l_2-n_2 l_1\right)}=\frac{n}{\left(l_1 m_2-l_2 m_1\right)}=\frac{\sqrt{l^2+m^2+n^2}}{\sqrt{\Sigma\left(m_1 n_2-m_2 n_1\right)^2}}\)

or \(\frac{l}{\left(m_1 n_2-m_2 n_1\right)}=\frac{m}{\left(n_1 l_2-n_2 l_1\right)}=\frac{n}{\left(l_1 m_2-l_2 m_1\right)}=\frac{1}{\sin \theta}\),

But, \(\theta=\frac{\pi}{2}\), and therefore, sin θ = 1.

∴ l = (m₁n₂ – m₂n₁), m = (n₁l₂ – n₂l₁) and n = (l₁m₂ – l₂m₁).

Hence, the direction cosines of the required line are (m₁n₂ – m₂n₁), (n₁l₂ – n₂l₁), (l₁m₂ – l₂m₁).

Shortest Distance between Two Lines

Coplanar Lines Two lines lying in the same plane are called coplanar lines. Coplanar lines are either parallel or intersecting.

Skew Lines Two lines in space which are not coplanar are called skew lines. Skew lines are neither parallel nor intersecting.

Line Of Shortest Distance Between Two Skew Lines If L₁ and L₂ are two skew lines then there is a unique line which is perpendicular to both the lines L₁ and L₂. This line is called the line of shortest distance between L₁ and L₂.

Shortest Distance Between Two Skew Lines The length of the line segment \(\overrightarrow{P Q}\), intercepted by two skew lines L₁ and L₂ on the common perpendicular to both the lines, is called the shortest distance (SD) between L₁ and L₂.

Remark If two lines in space intersect at a point then the shortest distance between them is zero.

To Find the Shortest Distance between Two Skew Lines

Vector Form

Theorem The shortest distance between two skew lines \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\) is given by \(d=\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right| .\)

Proof Let L₁ and L₂ be two skew lines whose vector equations are respectively.

\(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\) …(1)

and \(\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\) …(2)

Then, L₁ is parallel to \(\overrightarrow{b_1}\) and passes through a point A, whose position vector is \(\overrightarrow{a_1}\).

And, L₂ is parallel to \(\overrightarrow{b_2}\) and passes through a point B, whose position vector is \(\overrightarrow{a_2}\).

Let \(\overrightarrow{P Q}\) be the shortest-distance vector between L1 and L2.

Then, \(\overrightarrow{P Q} \perp \overrightarrow{b_1} \text { and } \overrightarrow{P Q} \perp \overrightarrow{b_2}\)

∴ \(\overrightarrow{P Q} \|\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\)

∴ \(P Q=\mid \text { projection of } \overrightarrow{A B} \text { along }\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \mid\)

= \(\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right| .\)

Condition For Two Given Lines To Intersect Suppose that the lines \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\) intersect. Then, the shortest distance between them is zero.

∴ \(\left[\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \overrightarrow{b_1} \overrightarrow{b_2}\right]=0\)

Remark Two lines intersect only when the shortest distance between them is zero.

Solved Examples

Example 1 Find the shortest distance between two lines whose vector equations are \(\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k}) \text {. }\)

Solution

Comparing the given equations with the standard equations \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\), we have

\(\overrightarrow{a_1}=(\hat{i}+2 \hat{j}+\hat{k}), \overrightarrow{b_1}=(\hat{i}-\hat{j}+\hat{k})\) \(\overrightarrow{a_2}=(2 \hat{i}-\hat{j}-\hat{k}) \text { and } \overrightarrow{b_2}=(2 \hat{i}+\hat{j}+2 \hat{k}) \text {. }\)

∴ \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(2 \hat{i}-\hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})=(\hat{i}-3 \hat{j}-2 \hat{k})\)

and, \(\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 1 \\
2 & 1 & 2
\end{array}\right|\)

= \((-2-1) \hat{i}-(2-2) \hat{j}+(1+2) \hat{k}\)

= \((-3 \hat{i}+3 \hat{k})\)

∴ \(\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{(-3)^2+3^2}=\sqrt{18}=3 \sqrt{2} .\)

∴ SD = \(\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right|\)

= \(\frac{|(-3 \hat{i}+3 \hat{k}) \cdot(\hat{i}-3 \hat{j}-2 \hat{k})|}{3 \sqrt{2}}=\frac{|(-3-0-6)|}{3 \sqrt{2}}\)

= \(\frac{|-9|}{3 \sqrt{2}}=\frac{9}{3 \sqrt{2}}=\frac{3 \sqrt{2}}{2} \text { units. }\)

The shortest distance between two lines = \(\frac{|-9|}{3 \sqrt{2}}=\frac{9}{3 \sqrt{2}}=\frac{3 \sqrt{2}}{2} \text { units. }\)

Example 2 Find the shortest distance between the lines L₁ and L₂ whose vector equations are given below:

L₁: \(\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k})\)

L₂: \(\vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})\)

Solution

Comparing the given equations with the standard equations \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\), we have

\(\overrightarrow{a_1}=(\hat{i}+\hat{j}), \overrightarrow{b_1}=(2 \hat{i}-\hat{j}+\hat{k})\) \(\overrightarrow{a_2}=(2 \hat{i}+\hat{j}-\hat{k}) \text { and } \overrightarrow{b_2}=(3 \hat{i}-5 \hat{j}+2 \hat{k})\)

∴ \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(2 \hat{i}+\hat{j}-\hat{k})-(\hat{i}+\hat{j})=(\hat{i}-\hat{k})\)

and, \(\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|\)

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

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

∴ \(\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{3^2+(-1)^2+(-7)^2}=\sqrt{59}\)

∴ SD = \(\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right|\)

= \(\frac{|(3 \hat{i}-\hat{j}-7 \hat{k}) \cdot(\hat{i}-\hat{k})|}{\sqrt{59}}=\frac{|3 \times 1-1 \times 0-7 \times(-1)|}{\sqrt{59}}\)

= \(\frac{|3-0+7|}{\sqrt{59}}=\frac{10 \sqrt{59}}{59} \text { units. }\)

The shortest distance between the lines L₁ and L₂  = \(\frac{|3-0+7|}{\sqrt{59}}=\frac{10 \sqrt{59}}{59} \text { units. }\)

Example 3 Find the shortest distance between the lines whose vector equations are

\(\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \text {, and }\) \(\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k} \text {. }\)

Solution

The given equations can be written as

\(\vec{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+t(-\hat{i}+\hat{j}-2 \hat{k}) \text {, and }\) \(\vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2 \hat{j}-2 \hat{k})\)

Comparing the given equations with the standard equations \(\vec{r}=\overrightarrow{a_1}+t \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+s \overrightarrow{b_2}\), we get

\(\overrightarrow{a_1}=(\hat{i}-2 \hat{j}+3 \hat{k}), \overrightarrow{b_1}=(-\hat{i}+\hat{j}-2 \hat{k})\) \(\vec{a}_2=(\hat{i}-\hat{j}-\hat{k}) \text { and } \vec{b}_2=(\hat{i}+2 \hat{j}-2 \hat{k}) \text {. }\)

∴ \(\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{2^2+(-4)^2+(-3)^2}=\sqrt{29}\)

∴ SD = \(\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right|\)

= \(\frac{|(2 \hat{i}-4 \hat{j}-3 \hat{k}) \cdot(\hat{j}-4 \hat{k})|}{\sqrt{29}}=\frac{|2 \times 0-4 \times 1-3 \times(-4)|}{\sqrt{29}}\)

= \(\frac{|0-4+12|}{\sqrt{29}}=\frac{8 \sqrt{29}}{29} \text { units. }\)

The shortest distance between the lines = \(\frac{|0-4+12|}{\sqrt{29}}=\frac{8 \sqrt{29}}{29} \text { units. }\)

Example 4 Show that the lines \(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j}) \text { and } \vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+3 \hat{k})\) intersect. Find their point of intersection.

Solution

Comparing the given equations with the standard equations \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\), we get

\(\overrightarrow{a_1}=(\hat{i}+\hat{j}-\hat{k}), \overrightarrow{b_1}=(3 \hat{i}-\hat{j})\) \(\overrightarrow{a_2}=(4 \hat{i}-\hat{k}) \text { and } \overrightarrow{b_2}=(2 \hat{i}+3 \hat{k}) \text {. }\)

∴ \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(4 \hat{i}-\hat{k})-(\hat{i}+\hat{j}-\hat{k})=(3 \hat{i}-\hat{j}) .\)

And, \(\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 0 \\
2 & 0 & 3
\end{array}\right|\)

= \((-3-0) \hat{i}-(9-0) \hat{j}+(0+2) \hat{k}\)

= \((-3 \hat{i}-9 \hat{j}+2 \hat{k})\)

∴ \(\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{(-3)^2+(-9)^2+2^2}=\sqrt{94}\)

∴ SD = \(\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right|\)

= \(\left|\frac{(-3 \hat{i}-9 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}-\hat{j})}{\sqrt{94}}\right|\)

= \(\frac{|-9+9+0|}{\sqrt{94}}=0\)

Thus, for some particular values of λ and μ, we have

\((\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j})=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+3 \hat{k})\)

⇒ \((1+3 \lambda) \hat{i}+(1-\lambda) \hat{j}-\hat{k}=(4+2 \mu) \hat{i}+(3 \mu-1) \hat{k}\)

⇒ 1 + 3λ = 4 + 2μ, 1 – λ = 0 and 3μ – 1 = -1

⇒ λ = 1 and μ = 0.

Thus, the position vector of the point of intersection of the given lines is given by

\(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+(3 \hat{i}-\hat{j}) [putting λ = 1], i.e., \vec{r}=(4 \hat{i}-\hat{k}) .\)

Hence, the point of intersection of the given lines if P(4, 0, -1).

Example 5 Show that the lines \(\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k}) \text { and } \vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}+\hat{j}-\hat{k})\) do not intersect.

Solution

Comparing the given equations with the standard equations \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\), we get

\(\overrightarrow{a_1}=(\hat{i}-\hat{j}), \overrightarrow{b_1}=(2 \hat{i}+\hat{k})\) \(\vec{a}_2=(2 \hat{i}-\hat{j}) \text { and } \vec{b}_2=(\hat{i}+\hat{j}-\hat{k})\)

∴ \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(2 \hat{i}-\hat{j})-(\hat{i}-\hat{j})=\hat{i} .\)

And, \(\left(\vec{b}_1 \times \vec{b}_2\right)=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 1 \\
1 & 1 & -1
\end{array}\right|\)

= \((0-1) \hat{i}-(-2-1) \hat{j}+(2-0) \hat{k}\)

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

∴ \(\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{(-1)^2+3^2+2^2}=\sqrt{14}\)

∴ SD = \(\left|\frac{\left(\vec{b}_1 \times \vec{b}_2\right) \cdot\left(\vec{a}_2-\vec{a}_1\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)

= \(\left|\frac{(-\hat{i}+3 \hat{j}+2 \hat{k}) \cdot \hat{i}}{\sqrt{14}}\right|=\frac{|-1|}{\sqrt{14}}\)

= \(\frac{1 \times \sqrt{14}}{14}=\frac{\sqrt{14}}{14} \neq 0\)

Since the shortest distance between the given lines is not zero, the given lines do not intersect.

Distance Between Parallel Lines Let L₁ and L₂ be two parallel lines. Then, these lines are clearly coplanar.

Let the equations of these lines be

\(\vec{r}=\overrightarrow{a_1}+\lambda \vec{b}\) …(1)

\(\vec{r}=\overrightarrow{a_2}+\mu \vec{b}\) …(2)

Let A be a point on L₁ with position vector \(\overrightarrow{a_1}\) and let B be a point on L₂ with position vector \(\overrightarrow{a_2}\).

Draw BM ⊥ L₁. Then, distance between L₁ and L₂ = \(|\overrightarrow{B M}|\).

Let θ be the angle between \(\overrightarrow{A B} \text { and } \vec{b}. Then, (\vec{b} \times \overrightarrow{A B})=\{|\vec{b}||\overrightarrow{A B}| \cdot \sin \theta\} \hat{n}\), where \(\hat{n}\) is a unit vector, perpendicular to the plane of L₁ and L₂.

∴ \(\vec{b} \times\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=|| \vec{b}|| \overrightarrow{A B}|\cdot \sin \theta| \hat{n}\)

⇒ \(\vec{b} \times\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=|\vec{b}|(B M) \vec{n}\) [∵ (AB) sin θ = BM]

⇒ \(\left|\vec{b} \times\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\right|=|\vec{b}||\overrightarrow{B M}| . 1\) [∵ \(|\hat{n}|=1\)]

⇒ \(|\overrightarrow{B M}|=\left|\frac{\vec{b} \times\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{|\vec{b}|}\right| .\)

Example 6 Find the shortest distance between the lines L1 and L2, given by \(\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(4 \hat{i}-2 \hat{j}+2 \hat{k})\).

Solution

The given lines are

L_1: \(\vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}-\hat{j}+\hat{k})\) …(1)

L_2: \(\vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+2 \mu(2 \hat{i}-\hat{j}+\hat{k})\) …(2)

These equations are of the form:

\(\vec{r}=\overrightarrow{a_1}+\lambda \vec{b} \text { and } \vec{r}=\overrightarrow{a_2}+2 \mu \vec{b}=\overrightarrow{a_2}+\mu^{\prime} \vec{b} \text {, where }\)

\(\overrightarrow{a_1}=(\hat{i}+\hat{j})\), \(\overrightarrow{a_2}=(2 \hat{i}+\hat{j}-\hat{k}), \vec{b}=(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \mu^{\prime}=2 \mu \text {. }\)

Clearly, the given lines are parallel.

Now, \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(2 \hat{i}+\hat{j}-\hat{k})-(\hat{i}+\hat{j})=(\hat{i}-\hat{k})\)

∴ \(\left\{\vec{b} \times\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\right\}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
1 & 0 & -1
\end{array}\right|\)

= \((1-0) \hat{i}-(-2-1) \hat{j}+(0+1) \hat{k}\)

= \((\hat{i}+3 \hat{j}+\hat{k})\)

⇒ \(\left|\vec{b} \times\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\right|=\sqrt{1^2+3^2+1^2}=\sqrt{11} and |\vec{b}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}\)

⇒ shortest distance between L₁ and L₂

= distance between L₁ and L₂

= \(\frac{\left|\vec{b} \times\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\right|}{|\vec{b}|}=\frac{\sqrt{11}}{\sqrt{6}}=\left(\frac{\sqrt{11}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}\right)=\frac{\sqrt{66}}{6} \text { units. }\)

Shortest Distance between Two Skew Lines in the Cartesian Form

The shortest distance between the skew lines \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \text { and } \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\) is given by

\(\mathrm{SD}=\frac{\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}{\sqrt{D}}\)

where D = {(a₁b₂ – a₂b₁)² + (b₁c₂ – b₂c₁)² + (c₁a₂ – c₂a₁)²}.

Condition For Two Given Lines To Intersect Let L₁ and L₂ be the given lines whose equations are

\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \text { and } \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)

1. L₁ and L₂ intersect ⇔ SD between them is 0

⇔ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0\)

2. L₁ and L₂ do not intersect ⇔ they are skew lines

⇔ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right| \neq 0 .\)

Solved Examples

Example 1 Find the shortest distance between the lines

\(\frac{x+3}{-4}=\frac{y-6}{3}=\frac{z}{2} \text { and } \frac{x+2}{-4}=\frac{y}{1}=\frac{z-7}{1} \text {. }\)

Solution

Comparing the given equations with

\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \text { and } \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\), we get

(x₁ = -3, y₁ = 6, z₁ = 0), (x₂ = -2, y₂ = 0, z₂ = 7), (a₁ = -4, b₁ = 3, c₁ = 2) and (a₂ = -4, b₂ = 1, c₂ = 1).

Now, D = (a₁b₂ – a₂b₁)² + (b₁c₂ – b₂c₁)² + (c₁a₂ – c₂a₁)²

= (-4+12)² + (3-2)² + (-8+4)²

= (64 + 1 + 16) = 81.

∴ SD = \(\frac{1}{\sqrt{D}} \cdot\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\)

= \(\frac{1}{\sqrt{81}} \cdot\left|\begin{array}{ccc}
-2+3 & 0-6 & 7-0 \\
-4 & 3 & 2 \\
-4 & 1 & 1
\end{array}\right|=\frac{1}{9} \cdot\left|\begin{array}{rrr}
1 & -6 & 7 \\
-4 & 3 & 2 \\
-4 & 1 & 1
\end{array}\right|\)

= \(\frac{1}{9} \cdot\{1 \cdot(3-2)+6 \cdot(-4+8)+7 \cdot(-4+12)\}=\frac{81}{9}=9 \text { units. }\)

Hence, the shortest distance between the given lines is 9 units.

Alternative Method

Example 2 Find the length and the equations of the line of shortest distance between the lines

\(\frac{x-3}{3}=\frac{y-8}{-1}=z-3 \text { and } \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4} \text {. }\)

Solution

The given equations are

\(\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}=\lambda \text { (say) }\) …(1)

\(\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}=\mu \text { (say) }\) …(2)

P(3λ+3, -λ+8, λ+3) is any point on (1).

Q(-3μ-3, 2μ-7, 4μ+6) is any point on (2).

The direction ratios of PQ are (-3μ – 3 – 6, 2μ + λ- 15, 4μ -λ + 3).

If PQ is the line of shortest distance then PQ is perpendicular to each of (1) and (2).

∴ \(\left\{\begin{array}{c}
3(-3 \mu-3 \lambda-6)-1 \cdot(2 \mu+\lambda-15)+1 \cdot(4 \mu-\lambda+3)=0 \\
\quad \text { and } \\
-3(-3 \mu-3 \lambda-6)+2 \cdot(2 \mu+\lambda-15)+4 \cdot(4 \mu-\lambda+3)=0
\end{array}\right.\)

⇒ \(\left\{\begin{array}{l}
-11 \lambda-7 \mu=0 \\
7 \lambda+29 \mu=0
\end{array}\right.\)

⇒ λ = 0 and μ = 0.

Thus, PQ will be the line of shortest distance with λ = 0 and μ = 0.

Substituting λ = 0 and μ = 0 in P and Q respectively, we get the points P(3, 8, 3) and Q(-3, -7, 6).

∴ SD = PQ = \(\sqrt{(-3-3)^2+(-7-8)^2+(6-3)^2}\)

= \(\sqrt{36+225+9}=\sqrt{270}=3 \sqrt{30} \text { units. }\)

Equations of the line of shortest distance means equations of PQ, given by

\(\frac{x-3}{-3-3}=\frac{y-8}{-7-8}=\frac{z-3}{6-3}\)

⇔ \(\frac{x-3}{-6}=\frac{y-8}{-15}=\frac{z-3}{3}\)

⇔ \(\frac{x-3}{2}=\frac{y-8}{5}=\frac{z-3}{-1}\)

Example 3 Find the length and the equations of the line of shortest distance between the lines

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \text { and } \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5} \text {. }\)

Solution

The given lines are

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda \text { (say) }\) …(1)

\(\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}=\mu \text { (say) }\) …(2)

P(2λ+1, 3λ+2, 4λ+3) is any point on (1).

Q(3μ+2, 4μ+4, 5μ+5) is any point on (2).

The direction ratios of \overrightarrow{P Q} are (3μ – 2λ + 1, 4μ – 3λ + 2, 5μ – 4λ + 2).

If \(\overrightarrow{P Q}\) is the line of shortest distance then \(\overrightarrow{P Q}\) is perpendicular to each of (1) and (2).

∴ \(\left\{\begin{array}{l}
2(3 \mu-2 \lambda+1)+3(4 \mu-3 \lambda+2)+4(5 \mu-4 \lambda+2)=0, \\
\quad \text { and } \\
3(3 \mu-2 \lambda+1)+4(4 \mu-3 \lambda+2)+5(5 \mu-4 \lambda+2)=0
\end{array}\right.\)

⇒ \(\left\{\begin{array}{l}
38 \mu-29 \lambda+16=0 \\
50 \mu-38 \lambda+21=0
\end{array}\right.\)

⇒ \(\frac{\mu}{(-609+608)}=\frac{\lambda}{(800-798)}=\frac{1}{(-1444+1450)}\) [by cross multiplication]

⇒ \(\frac{\mu}{-1}=\frac{\lambda}{2}=\frac{1}{6}\)

⇒ \(\lambda=\frac{2}{6}=\frac{1}{3} \text { and } \mu=\frac{-1}{6} \text {. }\)

Thus PQ will be the line of shortest distance when λ = \(\frac{1}{3}\) and μ = \(-\frac{1}{6}\). Putting λ = \(\frac{1}{3}\) and μ = \(-\frac{1}{6}\) in P and Q respectively, we get the points \(P\left(\frac{5}{3}, 3, \frac{13}{3}\right) \text { and } Q\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right) \text {. }\)

∴ SD = PQ = \(\sqrt{\left(\frac{3}{2}-\frac{5}{3}\right)^2+\left(\frac{10}{3}-3\right)^2+\left(\frac{25}{6}-\frac{13}{3}\right)^2}\)

= \(\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}}=\frac{1}{\sqrt{6}} \text { units }=\frac{\sqrt{6}}{6} \text { units. }\)

Hence, \(\mathrm{SD}=\frac{\sqrt{6}}{6} \text { units. }\)

Direction ratios of PQ are \(\left(\frac{3}{2}-\frac{5}{3}\right),\left(\frac{10}{3}-3\right),\left(\frac{25}{6}-\frac{13}{3}\right) \text {, } i.e., -\frac{1}{6}, \frac{1}{3}, \frac{-1}{6}, \text { i.e. }-1,2,-1 \text {. }\)

∴ the equations of the line of shortest distance are \(\frac{\left(x-\frac{5}{3}\right)}{-1}=\frac{y-3}{2}=\frac{\left(z-\frac{13}{3}\right)}{-1} .\)

Example 4 Show that the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \text { and } \frac{x-4}{5}=\frac{y-1}{2}=z\) intersect each other. Find their point of intersection.

Solution

The given lines are

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda \text { (say) }\) …(1)

\(\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}=\mu \text { (say) }\) …(2)

P(2λ+1, 3λ+2, 4λ+3) is any point on (1).

Q(5μ+4, 2μ+1, μ) is any point on (2).

If the lines (1) and (2) intersect then P and Q must coincide for some particular values of λ and μ.

This gives

2λ + 1 = 5μ + 4, 3λ + 2μ = 2 + 1 and 4λ + 3 = μ

⇔ 2λ – 5μ = 3 …(1), 3λ – 2μ = -1 …(2), and 4λ – μ = -3 …(3)

Solving (1) and (2), we get λ = -1 and μ = -1.

And, these values of λ and μ also satisfu (3).

Hence, the given lines intersect.

The point of intersection of the given lines is (-1, -1, -1), which is obtained by putting λ = -1 in P or μ = -1 in Q.

Example 5 Show that the lines \(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} \text { and } \frac{x-2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}\) do not intersect each other.

Solution

The given lines are

\(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda \text { (say) }\) …(1)

\(\frac{x-2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu \text { (say) }\) …(2)

P(3λ+1, 2λ-1, 5λ+1) is any point on (1).

Q(4μ+2, 3μ+1, -2μ-1) is any point on (2).

If the lines (1) and (2) intersect then P and Q must coincide for some particular values of λ and μ.

This gives

3λ + 1 = 4μ + 2, 2λ – 1 = 3μ + 1, 5λ + 1 = -2μ – 1

⇔ 3λ – 4μ = 1 …(1), 2λ – 3μ = 2 …(2), and 5λ + 2μ = -2 …(3).

Solving (1) and (2), we get λ = -5 and μ = -4.

But, the values λ = -5 and μ = -4 do not satisfy (3).

Hence, the given lines do not intersect each other.