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

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

hard 5 marks Class 12 · Mathematics · Three Dimensional Geometry

Answer

\frac{7}{3 10}

The correct answer is $\frac{7}{3 10}$ — Finally, the shortest distance $d = \frac{7}{3 10}$ .

Choose points on the lines: $P_{1} = (1, - 2, 5)$ from $L_{1}$ and $P_{2} = (- 3, 4, 2)$ from $L_{2}$ .
Determine direction vectors: $a_{1} = (2,, - 3,, 1)$ and $a_{2} = (1,, 2,, - 2)$ .
Compute the cross product $a_{1} \times a_{2} = (4,, 5,, 7)$ .
Find vector between the chosen points $P_{2} P_{1} = P_{2} - P_{1} = (- 4,, 6,, - 3)$ .
Calculate the scalar triple product $∣ P_{2} P_{1} \cdot (a_{1} \times a_{2}) ∣ = ∣ - 4 \cdot 4 + 6 \cdot 5 + (- 3) c d o t 7∣ = 7$ .
Determine the magnitude of the cross product $∣ a_{1} \times a_{2} ∣ = 4^{2} + 5^{2} + 7^{2} = 90 = 310$ .
Finally, the shortest distance $d = \frac{7}{3 10}$ .

Common mistake: Using wrong direction vectors or forgetting absolute value

Concept tested: Shortest distance between skew lines