JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Three Dimensional Geometry

The distance of the point having position vector

- \hat{i} + 2 \hat{j} + 6 \hat{k}

from the straight line passing through the point (

2, 3, - 4

) and parallel to the vector

6 \hat{i} + 3 \hat{j} - 4 \hat{k}

(A)7✓ Correct
(B) $43$
(C) $213$
(D)6

Answer: (A)

Solution

Let

P = (- 1, 2, 6)

and take the line through

A = (2, 3, - 4)

along

b = 6 \hat{i} + 3 \hat{j} - 4 \hat{k}

. Then

A P = (- 3, - 1, 10)

, and the distance is

\frac{∣ A P \times b ∣}{∣ b ∣}

. The cross product is

A P \times b = - 26 \hat{i} + 48 \hat{j} - 3 \hat{k}

, with magnitude

676 + 2304 + 9 = 2989

, while

∣ b ∣ = 36 + 9 + 16 = 61

. Dividing,

\frac{2989}{61} = 49 = 7

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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