JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Three Dimensional Geometry

The distance of the point

(1, - 2, 3)

from the plane

x - y + z = 5

measured parallel to a line, whose direction ratios are

2, 3, - 6

(A)2
(B)3
(C)1✓ Correct
(D)5

Answer: (C)

Solution

The line through

P (1, - 2, 3)

with direction ratios

(2, 3, - 6)

is:

\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} = k

A general point on this line is

Q (2 k + 1, 3 k - 2, - 6 k + 3)

.

Since

Q

lies on the plane

x - y + z = 5

(2 k + 1) - (3 k - 2) + (- 6 k + 3) = 5

2 k + 1 - 3 k + 2 - 6 k + 3 = 5

- 7 k + 6 = 5

k = \frac{1}{7}

Q (\frac{9}{7}, \frac{- 11}{7}, \frac{15}{7})

The required distance is:

P Q = (\frac{2}{7})^{2} + (\frac{3}{7})^{2} + (\frac{- 6}{7})^{2} = \frac{4 + 9 + 36}{49} = \frac{49}{49} = 1

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option

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