Question

Let d be the distance between the foot of perpendiculars of the point P(1, 2, –1) and Q(2, –1, 3) on the plane –x + y + z = 1. Then d² is equal to ____.

Answer 1

Correct Answer: 26

To find the squared distance d² between the foot of the perpendiculars of points P(1, 2, –1) and Q(2, –1, 3) on the plane –x + y + z = 1, we first need to determine the foot of the perpendicular for each point. The formula for the foot of the perpendicular (F) from a point (x₁, y₁, z₁) onto a plane Ax + By + Cz + D = 0 is:

(x₁ - Aλ, y₁ - Bλ, z₁ - Cλ),
where λ = (Ax₁ + By₁ + Cz₁ + D)/(A² + B² + C²).

The plane equation is –x + y + z = 1, or -1x + 1y + 1z - 1 = 0, giving A = -1, B = 1, C = 1, D = -1.
 

For P(1, 2, –1):
λ = [(-1)(1) + (1)(2) + (1)(-1) + (-1)]/((-1)² + 1² + 1²)
λ = (-1 + 2 - 1 - 1)/(1 + 1 + 1)
λ = -1/3

The foot of the perpendicular from P is:
(1 - (-1)(-1/3), 2 - 1(-1/3), -1 - 1(-1/3))
(1 + 1/3, 2 + 1/3, -1 + 1/3)
(4/3, 7/3, -2/3)

For Q(2, –1, 3):
λ = [(-1)(2) + (1)(-1) + (1)(3) + (-1)]/((-1)² + 1² + 1²)
λ = (-2 - 1 + 3 - 1)/(1 + 1 + 1)
λ = -1/3

The foot of the perpendicular from Q is:
(2 - (-1)(-1/3), -1 - 1(-1/3), 3 - 1(-1/3))
(2 + 1/3, -1 + 1/3, 3 + 1/3)
(7/3, -2/3, 10/3)

Now, calculate d² between these points:
d² = [(7/3) - (4/3)]² + [(-2/3) - (7/3)]² + [(10/3) - (-2/3)]²

d² = (3/3)² + (-9/3)² + (12/3)²
d² = 1² + 3² + 4²
d² = 1 + 9 + 16 = 26

Therefore, d² = 26, which matches the expected range of 26,26.