JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Three Dimensional Geometry

The distance of line

3 y - 2 z - 1 = 0 = 3 x - z + 4

from the point

(2, - 1, 6)

is:

(A) $26$ ✓ Correct
(B) $26$
(C) $25$
(D) $42$

Answer: (A)

Solution

The line is the intersection of

3 x - z + 4 = 0

and

3 y - 2 z - 1 = 0

. Taking

x = t

gives

z = 3 t + 4

and

y = 2 t + 3

, so the line passes through

(0, 3, 4)

with direction

d = (1, 2, 3)

. For

Q (2, - 1, 6)

, the vector from

(0, 3, 4)

Q

(2, - 4, 2)

, and

(2, - 4, 2) \times (1, 2, 3) = (- 16, - 4, 8) .

The distance is

\frac{∣ ( - 16 , - 4 , 8 ) ∣}{∣ ( 1 , 2 , 3 ) ∣} = \frac{336}{14} = 24 = 26

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option

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