JEE Main · 2017 · 02 Apr 2017 · Offline · Three Dimensional Geometry

The distance of the point

(1, 3, - 7)

from the plane passing through the point

(1, - 1, - 1)

, having normal perpendicular to both the lines

\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}

and

\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}

, is :

(A) $\frac{10}{74}$
(B) $\frac{20}{74}$
(C) $\frac{10}{83}$ ✓ Correct
(D) $\frac{5}{83}$

Answer: (C)

Solution

The plane's normal must be perpendicular to both lines, so take the cross product of their direction vectors

(1, - 2, 3)

and

(2, - 1, - 1)

, giving

n = (5, 7, 3)

. Through

(1, - 1, - 1)

the plane is

5 (x - 1) + 7 (y + 1) + 3 (z + 1) = 0

, that is

5 x + 7 y + 3 z + 5 = 0

. The distance of

(1, 3, - 7)

from it is

\frac{∣5 + 21 - 21 + 5∣}{25 + 49 + 9} = \frac{10}{83}

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option

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