Question

The equation of the plane containing the $\frac{x}{2} = \frac{y}{3} =\frac{z}{4}$ 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. x + 2y - 2z = 0
• B. x - 2y + z = 0
• C. 5x + 2y - 4z = 0
• D. 3x + 2y - 3z = 0

Answer 1

Answer: (B)

x - 2y + z = 0

Answer 2

Vector along the normal to the plane containing the lines
$\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$
is $\left(8\hat{i} -\hat{j} -10 \hat{k}\right)$
vector perpendicular to the vectors $2 \hat{i} + 3\hat{j} + 4\hat{k}$ and $8 \hat{i} - \hat{j} - 10\hat{k}$ is $26 \hat{i} - 52 \hat{j} + 26 \hat{k}$
so, required plane is $26x - 52y + 26z = 0$
$x - 2y + z = 0$