JEE Main 2025 Apr 2 Shift 2, Mathematics Q6: Lines in Space

JEE Main2025Apr 2, Shift 2Mathematics

The line $L_{1}$ is parallel to the vector $a = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$ and the line $L_{2}$ is parallel to the vector $b = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$ . The shortest distance between the lines $L_{1}$ and $L_{2}$ is :

A
$\frac{23}{38}$
B
$\frac{21}{57}$
C
$\frac{23}{57}$
D
$\frac{21}{38}$

Solution

Connecting vector between points: $P Q = (5 - 7, 3 - 6, 4 - 2) = (- 2, - 3, 2)$ .

Compute $a \times b$ :

$a \times b = \hat{i} \hat{j} \hat{k} - 324 213 = (6 - 4) \hat{i} - (- 9 - 8) \hat{j} + (- 3 - 4) \hat{k} = 2 \hat{i} + 17 \hat{j} - 7 \hat{k}$ .

$∣ a \times b ∣ = 4 + 289 + 49 = 342 = 338$ .

Shortest distance $= \frac{∣ P Q \cdot ( a \times b ) ∣}{∣ a \times b ∣} = \frac{∣ ( - 2 ) ( 2 ) + ( - 3 ) ( 17 ) + ( 2 ) ( - 7 ) ∣}{3 38} = \frac{∣ - 4 - 51 - 14∣}{3 38} = \frac{69}{3 38} = \frac{23}{38}$ .

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q6: Lines in Space

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