Question

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

Answer 1

Given: Equation of the plane is,

3x – y + 4z = 2   ...(i)

Let Q(x₁, y₁, z₁) be the image of the point P(3, –2, 1) in the given plane.

The equation of a line passing through (3, –2, 1) and perpendicular to the given plane is,

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

So, the coordinates of a general point on this line are (3k + 3, – k – 2, 4k + 1).

If N is the foot of the perpendicular from P to the given plane, then N lies on the plane.

Let the coordinates of N be (3k + 3, – k – 2, 4k + 1).

∴ 3(3k + 3) – (– k – 2) + 4(4k + 1) – 2 = 0   ...[Putting N in equation (i)]

⇒ 9k + 9 + k + 2 + 16k + 4 – 2 = 0

⇒ 26k + 13 = 0

⇒ `k = - 13/26`

= `- 1/2`

So, coordinates of N = `[3 xx (-1/2) + 3, -(-1/2) - 2, 4 xx (-1/2) + 1]`

= `(3/2, -3/2, -1)`

Since, N is the mid-point of PQ,

∴ By mid-point formula,

`(3 + x_1)/2 = 3/2`

⇒ x₁ = 3 – 3

⇒ x = 0

Also, `(-2 + y_1)/2 = -3/2`

⇒ y₁ = –3 + 2

⇒ y = –1

And `(1 + z_1)/2 = -1`

⇒ 1 + z₁ = –2

⇒ z₁ = –2 – 1

⇒ z = –3

Hence, the required image of the point P is Q(0, –1, –3).