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Question
Find the angle between the following pair of lines:
`vecr = 3hati + hatj - 2hatk + lambda(hati - hatj - 2hatk) and vecr = 2hati - hatj -56hatk + mu(3hati - 5hatj - 4hatk)`
Solution
Vectors of given measures are parallel to `b_1 = hati - hatj - 2hatk` and `b_2 = 3hati - 5hatj - 4hatk`, respectively.
∴ If the angle between these vectors is θ, then the angle between the lines will also be θ.
Then cos θ = `(vec(b_1). vec(b_2))/(|vec(b_1)|. |vec(b_2)|)`
= `((hati - hatj - 2hatk). (3hati - 5hatj - 4hatk))/(|hati - hatj - 2hatk|. |3hati - 5hatj - 4hatk|)`
= `((1)(3) + (-1)(-5) + (-2)(-4))/(sqrt(1 + 1 + 4). sqrt(9 + 25 + 16))`
= `(8sqrt3)/15`
⇒ θ = `cos^(-1) ((8sqrt3)/15)`
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