Default
Question
Find the shortest distance between the lines
→r = ˆi+ˆj+λ(2ˆi−ˆj+ˆk)
→r = 2ˆi+ˆj−ˆk+μ(3ˆi−5ˆj+2ˆk)
Solution
The correct answer is 10√59
Explanation
Here,
→a1 = ˆi+ˆj, →b1 = 2ˆi−ˆj+ˆk
→a2 = 2ˆi+ˆj−ˆk, →b2 = 3ˆi−5ˆj+2ˆk
∴ →a2 - →a1 = ˆi−ˆk
→b1 x →b2 =
|ˆi−ˆj−ˆk2−−1−13−−5−2| = 3ˆi−ˆj−7ˆk
∴ |→b1 x →b2| = √9+1+49 = √59
∴ The shortest distance between the given lines is
d = |(→b1x→b2).(→a2−→a1)|→b1x→b2||
= |3−0+7|√59
= 10√59
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