Let A be a point on the line $\overrightarrow{\mathrm{r}}=(1-3 \mu) \hat{\mathrm{i}}+(\mu-1) \hat{\mathrm{j}}+(2+5 \mu) \hat{\mathrm{k}}$ and $\mathrm{B}(3,2,6)$ be a point in the space. Then the value of $\mu$ for which the vector $\overrightarrow{\mathrm{AB}}$ is parallel to the plane $x-4 y+ 3 \mathrm{z}=1$ is:

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Three Dimensional Geometry

Let A be a point on the line

r = (1 - 3 μ) \hat{i} + (μ - 1) \hat{j} + (2 + 5 μ) \hat{k}

and

B (3, 2, 6)

be a point in the space. Then the value of

μ

for which the vector

AB

is parallel to the plane

x - 4 y + 3 z = 1

is:

(A) $\frac{1}{4}$ ✓ Correct
(B) $\frac{1}{8}$
(C) $\frac{1}{2}$
(D) $- \frac{1}{4}$

Answer: (A)

Solution

The point on the line is

A = (1 - 3 μ, μ - 1, 2 + 5 μ)

, so

A B = (2 + 3 μ, 3 - μ, 4 - 5 μ)

towards

B (3, 2, 6)

. For

A B

to be parallel to the plane

x - 4 y + 3 z = 1

it must be perpendicular to the normal

(1, - 4, 3)

, so

(2 + 3 μ) - 4 (3 - μ) + 3 (4 - 5 μ) = 0

. That reduces to

2 - 8 μ = 0

, giving

μ = \frac{1}{4}

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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