JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q14

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MathematicsSingle correct+4 / -1Question 14 of 90

If the two lines $l_1:\dfrac{x-2}3=\dfrac{y+1}{-2}$ , $z=2$ and $l_2:\dfrac{x-1}1=\dfrac{2y+3}\alpha=\dfrac{z+5}2$ are perpendicular, then an angle between the lines $l_2$ and $l_3:\dfrac{1-x}3=\dfrac{2y-1}{-4}=\dfrac z4$ is :

\cos^{-1}\left(\dfrac{29}4\right)

\sec^{-1}\left(\dfrac{29}4\right)

\cos^{-1}\left(\dfrac2{29}\right)

\cos^{-1}\left(\dfrac2{\sqrt{29}}\right)

Answer and worked solution

Official answer: B

Directions. $l_1$ : $(3,-2,0)$ . $l_2$ : $\left(1,\frac\alpha2,2\right)$ . Perpendicular: $3-\alpha=0$ , so $\alpha=3$ and $l_2\parallel(2,3,4)$ .

$l_3$ . Written in standard form its direction is $(-3,-2,4)$ .

Angle. $\cos\theta=\dfrac{-6-6+16}{\sqrt{29}\sqrt{29}}=\dfrac4{29}$ , i.e. $\theta=\sec^{-1}\dfrac{29}4$ .

(B)

Topics:Angles and distances Equation of a line in space

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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 14