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Question
Find the angle between the following pair of lines:
`vecr = 2hati - 5hatj + hatk + lambda(3hati - 2hatj + 6hatk) and vecr = 7hati - 6hatk + mu(hati + 2hatj + 2hatk)`
Solution
The vectors of the given equation are parallel to `b_1 = 3hati + 2hatj + 6hatk` and `b_2 = hati + 2hatj + 2hatk`, 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)|)`
= `((3hati + 2hatj + 6hatk). (hati + 2hatj + 2hatk))/(|3hati + 2hatj + 6hatk|. |hati + 2hatj + 2hatk|)`
= `(3 + 4 + 12)/(sqrt(3^2 + 2^2 + 6^2). sqrt(1^2 + 2^2 + 2^2))`
= `19/(sqrt49. sqrt9)`
= `19/(7 xx 3)`
= `19/21`
⇒ θ = `cos^(-1) 19/21`
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`vecr = 2hati - 5hatj + hatk + lambda(3hati + 2hatj + 6hatk)` and `vecr = 2hati - 5hatj + hatk + lambda(3hati + 2hatj + 6hatk)`
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The angle between two lines `(x + 1)/2 = (y + 3)/2 = (z - 4)/(-1)` and `(x - 4)/1 = (y + 4)/2 = (z + 1)/2` is ______.
A straight line L through the point (3, –2) is inclined at an angle of 60° to the line `sqrt(3)x + y` = 1. If L also intersects the x-axis, then the equation of L is ______.
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Find the angle between the following two lines:
`vecr = 2hati - 5hatj + hatk + λ(3hati + 2hatj + 6hatk)`
`vecr = 7hati - 6hatk + μ(hati + 2hatj + 2hatk)`