Consider the lines \[ \begin{aligned} & L_{1}: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2} \\ & L_{2}: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3} \end{aligned} \] The shortest distance between $L_{1}$ and $L_{2}$ is

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Three Dimensional Geometry

Consider the lines

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2} L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

The shortest distance between

L_{1}

and

L_{2}

(A)0
(B) $\frac{17}{3}$
(C) $\frac{41}{5 3}$
(D) $\frac{17}{5 3}$ ✓ Correct

Answer: (D)

Solution

The shortest distance between two skew lines is

\frac{∣ ( a _{2} - a _{1} ) \cdot ( d _{1} \times d _{2} ) ∣}{∣ d _{1} \times d _{2} ∣}

. Here

a_{1} = (- 1, - 2, - 1)

a_{2} = (2, - 2, 3)

, so

a_{2} - a_{1} = (3, 0, 4)

. From the previous question,

d_{1} \times d_{2} = (- 1, - 7, 5)

with magnitude

53

. The dot product is

(3, 0, 4) \cdot (- 1, - 7, 5) = - 3 + 0 + 20 = 17

. So the distance is

\frac{17}{5 3}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option

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