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JEE Main 2023 · 25 Jan · Shift 1

JEE Main 2023 Mathematics question on Three Dimensional Geometry

The distance of the point P (4, 6, –2) from the line passing through the point (–3, 2, 3) and parallel to a line with direction ratios 3, 3, –1 is equal to :
  1. A.
    232\sqrt{3}
  2. B.
    33
  3. C.
    14\sqrt{14}
  4. D.
    6\sqrt{6}
Show answer and solution

Correct option: C

Solution

Let A = (–3, 2, 3) and direction vector d = (3, 3, –1).

Vector AP = P – A = (4–(–3), 6–2, –2–3) = (7, 4, –5).

Compute AP × d:
|i j k |
|7 4 –5|
|3 3 –1|

= i(4·(–1) – (–5)·3) – j(7·(–1) – (–5)·3) + k(7·3 – 4·3)
= i(–4 + 15) – j(–7 + 15) + k(21 – 12)
= (11, –8, 9)

|AP × d| = √(121 + 64 + 81) = √266

|d| = √(9 + 9 + 1) = √19

Distance = √266 / √19 = √(266/19) = √14.

[JEE Main 2023 · 25 Jan · Shift 1]

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