Find the shortest distance between the skew lines L_1:(x-1)/2=(y+2)/-3=(z-4)/5 and L_2:(x+2)/-1=(y-3)/4=(z+1)/2.

Find the shortest distance between the skew lines $L_{1} : \frac{x - 1}{2} = \frac{y + 2}{- 3} = \frac{z - 4}{5}$ and $L_{2} : \frac{x + 2}{- 1} = \frac{y - 3}{4} = \frac{z + 1}{2}$ .

hard 4 marks Class 12 · Mathematics · Three Dimensional Geometry

Answer

\frac{18}{782}

The correct answer is $\frac{18}{782}$ — Distance $d = \frac{18}{782}$ .

Take a point on each line: $P_{1} = (1, - 2, 4)$ on $L_{1}$ and $P_{2} = (- 2, 3, 1)$ on $L_{2}$ .
Take direction vectors: $d_{1} = (2, - 3, 5)$ and $d_{2} = (- 1, 4, 2)$ .
Compute cross product $d_{1} \times d_{2}$ .
$d_{1} \times d_{2} = i 2 - 1 j - 3 4 k 52 = (- 26, - 9, 5)$ .
Compute magnitude $∣∣ d_{1} \times d_{2} ∣∣ = (- 26)^{2} + (- 9)^{2} + 5^{2} = 782$ .
Vector between the points: $P_{2} P_{1} = P_{2} - P_{1} = (- 3, 5, - 3)$ .
Dot this vector with the cross product: $(- 3) (- 26) + 5 (- 9) + (- 3) (5) = 78 - 45 - 15 = 18$ .
Absolute value $∣18∣ = 18$ is the numerator of distance formula.
Distance $d = \frac{18}{782}$ .

Common mistake: Using wrong sign or forgetting absolute value in numerator

Concept tested: Shortest distance between skew lines