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Question
Find the acute angle between the line `barr = (hati + 2hatj + 2hatk) + lambda(2hati + 3hatj - 6hatk)` and the plane `barr*(2hati - hatj + hatk)` = 0
Solution
The acute angle θ between the line `barr = bara + lambdabarb` and and the plane `barr*barn` = d is given by
sin θ = `|(barb*barn)/(|barb||barn|)|` ...(1)
Here, `barb = 2hati + 3hatj - 6hatk, barn = 2hati - hatj + hatk`
∴ `barb*barn = (2hati + 3hatj - 6hatk)*(2hati - hatj + hatk)`
= (2)(2) + (3)(– 1) + (– 6)(1)
= 4 – 3 – 6
= – 5
Also, `|barb| = sqrt(2^2 + 3^2 + (– 6)^2) = sqrt(49)` = 7
`|barn| = sqrt(2^2 + (– 1)^2 + 1^2) = sqrt(6)`
∴ From (1), we have,
sin θ = `|(-5)/(7sqrt(6))| = (5)/(7sqrt(6)`
∴ θ = `sin^-1(5/(7sqrt(6)))`
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