Question

Find the image of the point (–1, 3, 4) in the plane x – 2y = 0.

Answer 1

Equation of given plane be:

x – 2y + 0z = 0   ...(i)

Let Q be the foot of perpendicular from P(–1, 3, 4) to the plane (i).

∴ The equation of line PQ is

`(x + 1)/1 = (y - 3)/(-2) = (z - 4)/0 = λ`   ...(say)

For some values of λ, the co-ordinate of point Q are (λ – 1, –2λ + 3, 4)

Since, Q lies on the plane (i), then

(λ – 1) – 2 (–2λ + 3) = 0

λ – 1 + 4λ – 6 = 0

or 5λ = 7

or `λ = 7/5`

Co-ordinate of foot of perpendicular Q are `(7/5 - 1, -2(7/5) + 3, 4)`

or `Q(2/5, 1/5, 4)`

Let P'(x₁, y₁, z₁) be the image of P in the plane (i)

∴ Q is the mid-point of PP'.

∴ `(x_1 + (-1))/2 = 2/5`

`(3 + y_1)/2 = 1/5`

`(4 + z_1)/2 = 4`

or `x_1 = 9/5`

`y_1 = (-13)/5`

z₁ = 4

∴ Image of P(–1, 3, 4) in the plane (i) is `P'(9/5, (-13)/5, 4)`.