If the two lines l_1:(x-2)/(3)=(y+1)/(-2) , z=2 and l_2:(x-1)/(1)=(2y+3)/(alpha)=(z+5)/(2) are perpendicular, then an angle between the lines l_2 and l_3:(1-x)/(3)=(2y-1)/(-4)=(z)/(4) is :

3D Geometry: If the two lines l_1:(x-2)/(3)=(y+1)/(-2) , z=2 and… | JEE Main 2022

JEE Main 2022 (Online) 26th June Morning ShiftQuestion 1

If the two lines $l_1:\frac{x-2}{3}=\frac{y+1}{-2}$, $z=2$ and $l_2:\frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ are perpendicular, then an angle between the lines $l_2$ and $l_3:\frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is :

3D Geometry

$\cos^{-1}\left(\frac{29}{4}\right)$$\sec^{-1}\left(\frac{29}{4}\right)$✓ Correct answer$\cos^{-1}\left(\frac{2}{29}\right)$$\cos^{-1}\left(\frac{2}{\sqrt{29}}\right)$

Solution

$l_1$ direction $(3,-2,0)$; $l_2$ direction $(1,\alpha/2,2)$. Perpendicular: $3-\alpha=0\Rightarrow\alpha=3$, $l_2$: $(1,\tfrac32,2)\parallel(2,3,4)$. $l_3$ direction: $(-3,-2,4)$. $\cos\theta=\frac{|-6-6+16|}{\sqrt{29}\sqrt{29}}=\frac4{29}$, so $\theta=\sec^{-1}(29/4)$. Option (B).

3D Geometry - JEE Mathematics question (JEE Main 2022 (Online) 26th June Morning Shift)

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