Question

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

विकल्प

• 24

• 26

• 28

• 29

Answer 1

Let d be the distance between the foot of perpendiculars of the points P(1, 2, –1) and Q(2, –1, 3) on the plane –x + y + z = 1. Then d² is equal to 26.

Explanation:

Given: Equation of plane is –x + y + z = 1

Points P(1, 2, –1) and Q(2, –1, 3) lie on same side of the plane.

Now, perpendicular distance of a point p from plane – x + y + z – 1 = 0 is

 d₁ = `|(-1 + 2 - 1 - 1)/sqrt(1^2 + 1^2 + 1^2)| = 1/sqrt(3)`

And perpendicular distance of a point Q from plane –x + y + z₁ = 0 is

d₂ = `|(-2 - 1 + 3 - 1)/sqrt(1^2 + 1^2 + 1^2)| = 1/sqrt(3)`

∵ d₁ = d₂

∴ `vec(PQ)` is parallel to given plane.

So, distance between P and Q = distance between their foot of perpendiculars

⇒ `|vec(PQ)| = sqrt((2 - 1)^2 + (-1 - 2)^2 + (3 + 1)^2) = sqrt(26)`

⇒ d = `sqrt(26)`

⇒ d² = 26