Vectors & 3D GeometrymediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Vectors & 3D Geometry: Let Line Perpendicular Both Lines Acute Angle Between (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let a line LL be perpendicular to both the lines L1:x+13=y+35=z+57L_1:\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7} and L2:x−21=y−44=z−67L_2:\dfrac{x-2}{1}=\dfrac{y-4}{4}=\dfrac{z-6}{7}. If θ\theta is the acute angle between the lines LL and L3:x−8/72=y−4/71=z2L_3:\dfrac{x-8/7}{2}=\dfrac{y-4/7}{1}=\dfrac{z}{2}, then tan⁡θ\tan\theta is equal to
A322\dfrac32\sqrt2
B522\dfrac52\sqrt2correct
C532\dfrac53\sqrt2
D432\dfrac43\sqrt2
Solution
Step 1: d⃗1=(3,5,7)\vec d_1=(3,5,7), d⃗2=(1,4,7)\vec d_2=(1,4,7):
d⃗=d⃗1×d⃗2=∣i^j^k^357147∣.\vec d=\vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\3&5&7\\1&4&7\end{vmatrix}.
 i^:(5)(7)−(7)(4)=7\hat i:(5)(7)-(7)(4)=7; j^:−[(3)(7)−(7)(1)]=−14\hat j:-[(3)(7)-(7)(1)]=-14; k^:(3)(4)−(5)(1)=7\hat k:(3)(4)-(5)(1)=7.
d⃗=(7,−14,7)=7(1,−2,1).\vec d=(7,-14,7)=7(1,-2,1).
 Step 2: d⃗3=(2,1,2)\vec d_3=(2,1,2). (1,−2,1)⋅(2,1,2)=2−2+2=2(1,-2,1)\cdot(2,1,2)=2-2+2=2; ∣(1,−2,1)∣=6|(1,-2,1)|=\sqrt6, ∣(2,1,2)∣=3|(2,1,2)|=3:
cos⁡θ=236.\cos\theta=\dfrac{2}{3\sqrt6}.
 Step 3: cos⁡2θ=454⇒sin⁡θ=1−454=5054\cos^2\theta=\dfrac{4}{54}\Rightarrow\sin\theta=\sqrt{1-\dfrac{4}{54}}=\sqrt{\dfrac{50}{54}}.
tan⁡θ=50/542/(36)=5054⋅362=5054⋅542=502=522.\tan\theta=\dfrac{\sqrt{50/54}}{2/(3\sqrt6)}=\sqrt{\dfrac{50}{54}}\cdot\dfrac{3\sqrt6}{2}=\sqrt{\dfrac{50}{54}}\cdot\dfrac{\sqrt{54}}{2}=\dfrac{\sqrt{50}}{2}=\dfrac{5\sqrt2}{2}.
 Correct answer: (2)
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