If the two lines $l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ perpendicular, then an angle between the lines $l_{2}$ and $l_{3}: \frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is:

(B) $\sec^{-1}\left(\frac{29}{4}\right)$

(A) $\cos^{-1}\left(\frac{29}{4}\right)$

(C) $\cos^{-1}\left(\frac{2}{29}\right)$

(D) $\cos^{-1}\left(\frac{2}{\sqrt{29}}\right)$

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Three Dimensional Geometry

If the two lines

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

and

l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}

perpendicular, then an angle between the lines

l_{2}

and

l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4}

is:

Answer: (B)

Rewrite

l_{1}

in standard form:

\frac{x - 2}{3} = \frac{y + 1}{- 2} = \frac{z - 2}{0}

. For

l_{2}

\frac{x - 1}{1} = \frac{y + 3/2}{α /2} = \frac{z + 5}{2}

. Since

l_{1} ⊥ l_{2}

, their direction vectors are perpendicular:

3 (1) + (- 2) (α /2) + 0 (2) = 0

, giving

α = 3

. For

l_{3}

\frac{x - 1}{- 3} = \frac{y - 1/2}{- 2} = \frac{z}{4}

. The angle between

l_{2}

and

l_{3}

is found using

cos θ = \frac{∣1 ( - 3 ) + ( 3/2 ) ( - 2 ) + 2 ( 4 ) ∣}{1 + 9/4 + 4 9 + 4 + 16} = \frac{2}{29/4 29} = \frac{4}{29}

. Thus

θ = cos^{- 1} (\frac{4}{29}) = sec^{- 1} (\frac{29}{4})

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option