JEE Main 2025 Apr 4 Shift 2, Mathematics Q19: Lines in Space

JEE Main2025Apr 4, Shift 2Mathematics

Let $A$ be the point of intersection of the lines $L_{1} : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{- 1}$ and $L_{2} : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}$ . Let $B$ and $C$ be the points on the lines $L_{1}$ and $L_{2}$ respectively such that $A B = A C = 15$ . Then the square of the area of the triangle $A B C$ is :

A
54
B
63
C
57
D
60

Solution

The triangle is isosceles with $A B = A C = 15$ and the angle at $A$ equals the angle between the two lines.

Direction vectors: $d_{1} = (1, 0, - 1)$ with $∣ d_{1} ∣ = 2$ , and $d_{2} = (3, 4, 5)$ with $∣ d_{2} ∣ = 50$ .

$cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣ ∣ d _{2} ∣} = \frac{∣3 + 0 - 5∣}{2 50} = \frac{2}{10} = \frac{1}{5}$ .

So $sin θ = \frac{24}{5} = \frac{2 6}{5}$ .

Area $= \frac{1}{2} A B \cdot A C sin θ = \frac{1}{2} \cdot 15 \cdot \frac{2 6}{5} = 36$ .

$∴ (Area)^{2} = 9 \cdot 6 = 54$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q19: Lines in Space

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