Consider the lines \[ \begin{aligned} & L_{1}: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2} \\ & L_{2}: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3} \end{aligned} \] The distance of the point $(1,1,1)$ from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Three Dimensional Geometry

Consider the lines

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2} L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

The distance of the point

(1, 1, 1)

from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines

L_{1}

and

L_{2}

(A) $\frac{2}{75}$
(B) $\frac{7}{75}$
(C) $\frac{13}{75}$ ✓ Correct
(D) $\frac{23}{75}$

Answer: (C)

Solution

The normal to the plane is

n = d_{1} \times d_{2} = (- 1, - 7, 5)

. The plane passing through

(- 1, - 2, - 1)

with this normal is

- (x + 1) - 7 (y + 2) + 5 (z + 1) = 0

, which simplifies to

x + 7 y - 5 z + 10 = 0

. The distance from

(1, 1, 1)

to this plane is

\frac{∣1 + 7 - 5 + 10∣}{1 + 49 + 25} = \frac{13}{75}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option

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