The equation of the plane containing the straight line and perpendicular to the plane containing the straight lines is

Engineering

Mathematics

Plane and Its Different Forms

Distance from a Plane

Equation of Straight Line in 3D and Distance Of a Point From Line

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

5x + 2y – 4z = 0

x + 2y – 2z = 0

3x + 2y – 3z = 0

x – 2y + z = 0

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Vector along the normal to the plane containing the lines

$\frac{x}{3} = \frac{y}{4} = \frac{z}{2} and \frac{x}{4} = \frac{y}{2} = \frac{z}{3} is (8 \hat{i} - \hat{j} - 10 \hat{k}) .$

Vector perpendicular to the vectors $2 \hat{i} + 3 \hat{j} + 4 \hat{k} and 8 \hat{i} - \hat{j} - 10 \hat{k} - 52 \hat{j} + 26 \hat{k}$

So, required plane is 26x – 52y + 26z + 0 ⇒ x – 2y + z = 0