JEE Main 2026 Apr 6 Shift 1, Mathematics Q16: Lines in Space

JEE Main2026Apr 6, Shift 1Mathematics

Let a line $L$ be perpendicular to both the lines $L_{1} : \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $L_{2} : \frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}$ .

If $θ$ is the acute angle between the lines $L$ and $L_{3} : \frac{x - 8/7}{2} = \frac{y - 4/7}{1} = \frac{z}{2}$ , then $tan θ$ is equal to:

A
$\frac{3}{2} 2$
B
$\frac{5}{2} 2$
C
$\frac{5}{3} 2$
D
$\frac{4}{3} 2$

Solution

The direction of $L$ is the cross product of the directions of $L_{1}$ and $L_{2}$ :

$(3, 5, 7) \times (1, 4, 7) = (5 \cdot 7 - 7 \cdot 4, 7 \cdot 1 - 3 \cdot 7, 3 \cdot 4 - 5 \cdot 1) = (7, - 14, 7)$ , parallel to $(1, - 2, 1)$ .

Direction of $L_{3}$ is $(2, 1, 2)$ . Compute $cos θ$ :

$cos θ = \frac{∣1 \cdot 2 + ( - 2 ) \cdot 1 + 1 \cdot 2∣}{6 9} = \frac{2}{3 6}$ .

Then $sin θ = 1 - \frac{4}{54} = \frac{50}{54} = \frac{5 2}{54}$ , and

$tan θ = \frac{sin θ}{cos θ} = \frac{5 2 / 54}{2/ ( 3 6 )} = \frac{5 2}{2}$ .

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q16: Lines in Space

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