Find the shortest distance between the following pair of lines.
$¯ r = (¯ i + 2 ¯ j + ¯ k) + λ (2 ¯ i - ¯ j + 3 ¯ k)$ & $¯ r = (¯ i - 3 ¯ j - ¯ k) + μ (3 ¯ i + 2^j - 5 ¯ k)$ .

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Solution

$\begin{matrix} A c c o r d i n g t o q u e s t i o n, w e h a v e t w o p a i r o f l i n e . . . . . . . . . . {¯ r}_{1} = (¯ i + 2 ¯ j + ¯ k) + λ (2 ¯ i - ¯ j + 3 ¯ k) {¯ r}_{2} = (¯ i - 3 ¯ j - ¯ k) + μ (3 ¯ i + 2^j - 5 ¯ k) N o w, \end{matrix}$
$C o m p a r i n g t h e g i v e n e q u^{n} w i t h t h e e q u a t i o n s : {\to r}_{1} = {\to a}_{1} + λ \to b_{1} a n d \to r_{2} = {\to a}_{2} + λ \to b_{2}$
$\begin{matrix} {\to a}_{1} = ˆ i - 2 ˆ j + ˆ k {\to a}_{2} = ˆ i - 3 ˆ j - ˆ k \to b_{1} = 2 ˆ i - ˆ j + 3 ˆ k \to b_{2} = 3 ˆ i + 2 ˆ j - 5 ˆ k ∴ {- \to a}_{2} - {\to a}_{1} = (ˆ i - 3 ˆ j - ˆ k) - (ˆ i - 2 ˆ j + ˆ k) = ˆ i - 3 ˆ j - ˆ k - ˆ i + 2 ˆ j - ˆ k = (- ˆ j - 2 ˆ k) - - - - - (i) N o w \to b_{1} \times \to b_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} i ˆ ˆ j ˆ k 2 - 13 32 - 5 \end{matrix} ∣ ∣ ∣ ∣ ∣ = ˆ i (5 - 6) - ˆ j (- 10 - 9) + ˆ k (4 + 3) = - ˆ i + 19 ˆ j + 7 ˆ k - - - - - (i i) ∣ ∣ ∣ \to b_{1} \times \to b_{2} ∣ ∣ ∣ = \sqrt{{(- 1)}^{2} + {(19)}^{2} + {(7)}^{2}} = \sqrt{1 + 361 + 49} = \sqrt{411} = 21 ({- \to a}_{2} - {\to a}_{1}) (\to b_{1} \times \to b_{2}) = (- ˆ j - 2 ˆ k) (- ˆ i + 19 ˆ j + 7 ˆ k) = 1 + 19 - 14 = 6 a n d s h o r t e s t d i s tan c e b e t w e e n t h e l i n e : d = ∣ ∣ ∣ ∣ \frac{({- \to a}_{2} - {\to a}_{1}) (\to b_{1} \times \to b_{2})}{∣ ∣ ∣ \to b_{1} \times \to b_{2} ∣ ∣ ∣} ∣ ∣ ∣ ∣ = \frac{6}{21} = \frac{2}{7} s o t h a t t h e c o r r e c t s o l u t i o n \frac{2}{7} \end{matrix}$

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