Question

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

• A. $x+2y-2z=0$
• B. $3x+2y-3z=0$
• C. $x-2y+z=0$
• D. $5x+2y-4z=0$

Answer 1

The correct option is C $x-2y+z=0$

Let the plane be $P_1$ containing the straight lines
$\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\text{and}\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$

$\therefore$ Vector normal to the plane $P_1$ is $\overrightarrow{n_1}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 2\\ 4& 2& 3\end{array}\right|$

$=8\hat{i}-\hat{j}-10\hat{k}$

Let the required plane be $P_2$ which contains the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$
As plane $P_2$ is perpendicular to plane $P_1$
So, normal vector to the plane $P_2$ is $\overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 8& -1& -10\\ 2& 3& 4\end{array}\right|$

$\Rightarrow \overrightarrow{n_2}=26\hat{i}-52\hat{j}+26\hat{k}$

Plane $P_2$ passes through $(0,0,0)$
So, equation of plane $P_2$ is $26x-52y+26z=0$
$\Rightarrow x-2y+z=0$