JEE Main · 2016 · 03 Apr 2016 · Offline · Three Dimensional Geometry

The distance of the point

(1, - 5, 9)

from the plane

x - y + z = 5

measured along the line

x = y = z

is:

(A) $103$ ✓ Correct
(B) $\frac{10}{3}$
(C) $\frac{20}{3}$
(D) $310$

Answer: (A)

Solution

First take the ordinary perpendicular distance from

A (1, - 5, 9)

to the plane

x - y + z = 5

\frac{∣1 - ( - 5 ) + 9 - 5∣}{1 + 1 + 1} = \frac{10}{3}

. But the distance is measured along the line

x = y = z

, direction

(1, 1, 1)

, tilted from the plane's normal

(1, - 1, 1)

cos θ = \frac{1 - 1 + 1}{3 3} = \frac{1}{3}

. The slant distance is the perpendicular distance divided by this cosine,

\frac{10/ 3}{1/3} = \frac{30}{3} = 103

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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