3D Geometry Ques 32
32. 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:
Formula:
- 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} $
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