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

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

hard 5 marks Class 12 · Mathematics · Three Dimensional Geometry

Answer

\frac{4}{5}

The correct answer is $\frac{4}{5}$ — Step 6: Shortest distance $= \frac{4}{5} \approx 1.79$ units.

Step 1: Identify direction vectors: $d_{1} = (4, - 5, 2)$ from $L_{1}$ and $d_{2} = (- 2, 3, - 1)$ from $L_{2}$ .
Step 2: Choose a point on each line; $A (3, - 2, 1)$ on $L_{1}$ ( $t = 0$ ) and $B (- 1, 4, - 3)$ on $L_{2}$ ( $s = 0$ ).
Step 3: Form vector $AB = B - A = (- 4, 6, - 4)$ .
Step 4: Compute cross product $d_{1} \times d_{2} = (- 1, 0, 2)$ and its magnitude $∣ d_{1} \times d_{2} ∣ = (- 1)^{2} + 0^{2} + 2^{2} = 5$ .
Step 5: Compute triple scalar product $∣ AB \cdot (d_{1} \times d_{2}) ∣ = ∣ (- 4, 6, - 4) \cdot (- 1, 0, 2) ∣ = ∣4 - 8∣ = 4$ .
Step 6: Shortest distance $= \frac{4}{5} \approx 1.79$ units.

Common mistake: Using the difference of direction vectors instead of the cross product

Concept tested: Shortest distance between skew lines