Question

Find the acute angle between the line `barr = (hati + 2hatj + 2hatk) + lambda(2hati + 3hatj - 6hatk)` and the plane `barr*(2hati - hatj + hatk)` = 0

Answer 1

The acute angle θ between the line `barr = bara + lambdabarb` 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)))`