Find the shortest distance between the skew lines $L_{1} :; x = 1 + 3 t,; y = - 2 + 2 t,; z = 5 + t$ and $L_{2} :; (x - 4) /2 = (y + 3) / - 1 = (z - 1) /4$ .

hard 5 marks Class 12 · Mathematics · Three Dimensional Geometry

Answer

\frac{65}{230}

The correct answer is $\frac{65}{230}$ — Divide the absolute value of the scalar triple product by the magnitude of the cross product to obtain the distance $\frac{65}{230}$ .

Step-by-step solution

Write the direction vectors: for $L_{1}$ $d_{1} = (3, 2, 1)$ and for $L_{2}$ $d_{2} = (2, - 1, 4)$ .
Compute the cross product $d_{1} \times d_{2} = (9, - 10, - 7)$ .
Select a point on each line: $P (1, - 2, 5)$ on $L_{1}$ and $Q (4, - 3, 1)$ on $L_{2}$ .
Form the connecting vector $P Q = Q - P = (3, - 1, - 4)$ .
Calculate the scalar triple product $P Q \cdot (d_{1} \times d_{2}) = 65$ .
Find the magnitude of the cross product: $∣ d_{1} \times d_{2} ∣ = 9^{2} + (- 10)^{2} + (- 7)^{2} = 230$ .
Divide the absolute value of the scalar triple product by the magnitude of the cross product to obtain the distance $\frac{65}{230}$ .

Common mistake: Neglecting the absolute value in the scalar triple product or miscomputing the cross product.

Concept tested: Shortest distance between skew lines

Published 6 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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Find the shortest distance between the skew lines L1​:;x=1+3t,;y=−2+2t,;z=5+t and L2​:;(x−4)/2=(y+3)/−1=(z−1)/4.

Step-by-step solution

Find the shortest distance between the skew lines $L_{1} :; x = 1 + 3 t,; y = - 2 + 2 t,; z = 5 + t$ and $L_{2} :; (x - 4) /2 = (y + 3) / - 1 = (z - 1) /4$ .