Question

Find the point of intersection of the line with the plane (x – 1) = `y/2` = z + 1 with the plane 2x – y – 2z = 2. Also, the angle between the line and the plane

Answer 1

Any point on the line x – 1 = `y/2` = z + 1 is

x – 1 = `y/2` = z + 1 = λ, (say)

(λ + 1, 2λ, λ – 1)

This passes through the plane 2x – y + 2z = 2

2(λ + 1) – 2λ + 2(λ – 1) = 2

2λ + 2 – 2λ + 2λ – 2 = 2

λ = 1

∴ The required point of intersection is (2, 2, 0)

sin θ = `(|vec"b"*vec"n"|)/(|vec"b"||vec"n"|)`

`vec"b" = hat"i" + 2hat"j" + hat"k"`

`vec"n" = 2hat"i" - hat"j" + 2hat"k"`

`vec"b"*vec"n" = 2 - 2 + 2` = 2

`|vec"b"| = sqrt(1 + 4 + 1) = sqrt(6)`

`|vec"n"| = sqrt(4 + 1 + 1) = sqrt(9)` = 3

sin θ = `|2|/(sqrt(6) xx 3)`

= `2/(3sqrt(6))`

θ = `sin^-1(2/(3sqrt(6)))`