If an angle between the line, $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ and the plane, $x-2 y-k z=3$ is $\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$, then a value of k is :

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Three Dimensional Geometry

If an angle between the line,

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

and the plane,

x - 2 y - k z = 3

cos^{- 1} (\frac{2 2}{3})

, then a value of k is :

(A) $\frac{5}{3}$ ✓ Correct
(B) $\frac{3}{5}$
(C) $- \frac{3}{5}$
(D) $- \frac{5}{3}$

Answer: (A)

Solution

The line has direction ratios

(2, 1, - 2)

and the plane's normal is

(1, - 2, - k)

. The line

-

plane angle satisfies

sin α = \frac{∣2 - 2 + 2 k ∣}{9 5 + k ^{2}} = \frac{∣2 k ∣}{3 5 + k ^{2}}

. Given

cos α = \frac{2 2}{3}

sin α = \frac{1}{3}

, so

\frac{2 k}{3 5 + k ^{2}} = \frac{1}{3}

, i.e.

2 k = 5 + k^{2}

. Squaring,

4 k^{2} = 5 + k^{2}

, so

k^{2} = \frac{5}{3}

and

k = \frac{5}{3}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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