If the line drawn from the point (–2, – 1, – 3) meets a plane at

1.	If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.
Answer» Given, points (–2, – 1, – 3) and (1, – 3, 3) Direction ratios of the normal to the plane are (1 + 2, -3 + 1, 3 + 3) = (3, -2, 6) Now, the equation of plane passing through one point (x₁, y₁, z₁) is a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 3(x – 1) – 2(y + 3) + 6(z – 3) = 0 3x – 3 – 2y – 6 + 6z – 18 = 0 3x – 2y + 6z – 27 = 0 ⇒ 3x – 2y + 6z = 27 Thus, the required equation of plane is 3x – 2y + 6z = 27.

If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.

Answer»

Given, points (–2, – 1, – 3) and (1, – 3, 3)

Direction ratios of the normal to the plane are (1 + 2, -3 + 1, 3 + 3) = (3, -2, 6)

Now, the equation of plane passing through one point (x₁, y₁, z₁) is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

3(x – 1) – 2(y + 3) + 6(z – 3) = 0

3x – 3 – 2y – 6 + 6z – 18 = 0

3x – 2y + 6z – 27 = 0 ⇒ 3x – 2y + 6z = 27

Thus, the required equation of plane is 3x – 2y + 6z = 27.