Question

Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3 and at a distance `2/sqrt(3)` from the point (3, 1, –1)

Answer 1

Required equation of the plane

(x + 2y + 3z – 2) + λ(x – y + z – 3) = 0 ........(1)

(1 + λ)x + (2 – λ)y + (3 + λ)z + (– 2 – 3λ) = 0 ........(2)

Distance from (2) to the point (3, 1, –1) is `2/sqrt(3)`

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

`vec"n" = 6hat"i" + 3hat"j" + 2hat"k"`

`vec"b"*vec"n"` = 6 + 6 – 4 = 8

`|vec"b"| = sqrt(1 + 4 + 4) = sqrt(9)` = 0

`|vec"n"| = sqrt(36 + 9 + 4)` = `sqrt(49)` = 7

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

= `/(3 xx 7)`

= `8/21`

θ = `sin^-1(8/21)`

Putting λ = `(-7)/2` in (1)

The required equation

(x + 2y + 3z – 2) – `7/2` (x – y + z – 3) = 0

2x + 4y + 6z – 4 – 7x + 7y – 7z + 21 = 0

– 5x + 11y – z + 17 = 0

5x – 11y + z – 17 = 0