JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Three Dimensional Geometry

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}

Answer: (C)

Solution

The first line has direction vector

d_{1} = (2, 3, 4)

.

The plane containing the second and third lines has normal vector:

n_{2} = (3, 4, 2) \times (4, 2, 3) = \hat{i} 34 \hat{j} 42 \hat{k} 23 = (8, - 1, - 10)

The required plane contains

d_{1}

and is perpendicular to the plane with normal

n_{2}

, so its normal is:

n = d_{1} \times n_{2} = (2, 3, 4) \times (8, - 1, - 10) = (- 26, 52, - 26) = - 26 (1, - 2, 1)

The plane passes through the origin (the first line passes through origin), so the equation is:

x - 2 y + z = 0

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option