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Question
Find the acute angle between the following lines.
2x = 3y = – z and 6x = – y = – 4z
Solution
2x = 3y = – z
`x/(1/2) = y/(1/3) = z/(-1)`
`|vec"b"| = 1/2 vec"i" + 1/3vec"j" - vec"k"`
6x = – y = – 4z
`x/(1/6) = y/(-1) = z/((-1)/4)`
`|vec"d"| = 1/6 vec"i" - vec"j" - 1/4vec"k"`
`vec"b"*vec"d" = 1/12 - 1/3 + 1/4`
= `(1 - 4 + 3)/12`
= 0
cos θ = `|vec"b" * vec"d"|/(|vec"b"| |vec"d"|)`
= `0/(sqrt((1/2)^2 + (1/3)^2 + (- 1)^2) sqrt((1/6)^2 + (- 1)^2 + ((-1)/4)^2)`
⇒ θ = `cos^-1(0)`
cos θ = 0
θ = `pi/2`
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