3D Geometry 3 Question 26

####26. The equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

(a) $5 x+2 y-4 z=0$

(b) $x+2 y-2 z=0$

(c) $3 x+2 y-3 z=0$

(d) $x-2 y+z=0$

Show Answer

Answer:

(d)

Solution:

  1. Let $P_{1}$ be the plane containing the lines

$ \frac{x}{3}=\frac{y}{4}=\frac{z}{2} \text { and } \quad \frac{x}{4}=\frac{y}{2}=\frac{z}{3} $

For these two lines, direction vectors are

$ \mathbf{b}_ {1}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } \mathbf{b}_ {2}=4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} $

A vector along the normal to the plane $P_ {1}$ is given by

$ \begin{aligned} \mathbf{n} _ {1}=\mathbf{b}_ {1} \times \mathbf{b}_ {2} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{array}\right| \\ & =\hat{\mathbf{i}} (12- 4)- \hat{\mathbf{j}}(9- 8)+\hat{\mathbf{k}}(6-16)= 8 \hat{\mathbf{i}}- \hat{\mathbf{j}}- 10 \hat{\mathbf{k}} \end{aligned} $

Let $P_{2}$ be the plane containing the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to plane $P_{1}$.

For the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$, the direction vector is $\mathbf{b}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and it passes through the point with position vector $\mathbf{a}=0 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+0 \hat{\mathbf{k}}$.

$\because P_{2}$ is perpendicular to $P_{1}$, therefore $\mathbf{n}_{1}$ and $\mathbf{b}$ lies along the plane.

Also, $P_{2}$ also passes through the point with position vector $\mathbf{a}$.

$\therefore$ Equation of plane $P_{2}$ is given by

$ \begin{aligned} & (\mathbf{r}-\mathbf{a}) \cdot\left(\mathbf{n}_{1} \times \mathbf{b}\right)=0 \Rightarrow\left|\begin{array}{ccc} x-0 & y-0 & z-0 \\ 8 & -1 & -10 \\ 2 & 3 & 4 \end{array}\right|=0 \\ & \Rightarrow x(-4+30)-y(32+20)+z(24+2)=0 \\ & \Rightarrow 26 x-52 y+26 z=0 \\ & \Rightarrow \quad x-2 y+z=0 \end{aligned} $



NCERT Chapter Video Solution

Dual Pane