JEE Main 2026 · 6 Apr · Shift 1 · Mathematics Q16

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MathematicsSingle correct+4 / -1Question 16 of 75

Let a line L be perpendicular to both the lines $L_1:\dfrac{x+1}3=\dfrac{y+3}5=\dfrac{z+5}7$ and $L_2:\dfrac{x-2}1=\dfrac{y-4}4=\dfrac{z-6}7$ . If $\theta$ is the acute angle between the lines L and $L_3:\dfrac{x-\frac87}2=\dfrac{y-\frac47}1=\dfrac z2$ , then $\tan\theta$ is equal to:

\dfrac32\sqrt2

\dfrac52\sqrt2

\dfrac53\sqrt2

\dfrac43\sqrt2

Answer and worked solution

Official answer: B

Direction of L.
$(3,5,7)\times(1,4,7)=(7,-14,7)\parallel(1,-2,1).$

Angle with $L_3$ (direction $(2,1,2)$ ).
$\cos\theta=\frac{|2-2+2|}{\sqrt6\cdot3}=\frac2{3\sqrt6},\qquad\sin\theta=\sqrt{\frac{50}{54}}.$

So $\tan\theta=\dfrac{\sqrt{50}}2=\dfrac52\sqrt2$ .

(B)

Topics:Direction cosines and ratios Cross product

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JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 16