Let d be the distance between the foot of perpendiculars of the points $\mathrm{P}(1,2,-1)$ and $\mathrm{Q}(2,-1,3)$ on the plane $-x+y+z=1$. Then $d^{2}$ is equal to

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Three Dimensional Geometry

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

d^{2}

is equal to

Answer: 26

Solution

The perpendicular distance from

P (1, 2, - 1)

to the plane

- x + y + z = 1

is:

\frac{∣ - 1 + 2 - 1 - 1∣}{3} = \frac{1}{3}

The perpendicular distance from

Q (2, - 1, 3)

to the plane is:

\frac{∣ - 2 - 1 + 3 - 1∣}{3} = \frac{1}{3}

Since both points are at the same distance from the plane and on the same side, the line

P Q

is parallel to the plane.

Therefore, the distance between the feet of perpendiculars equals

∣ P Q ∣

d = ∣ P Q ∣ = (2 - 1)^{2} + (- 1 - 2)^{2} + (3 - (- 1))^{2} = 1 + 9 + 16 = 26

Thus

d^{2} = 26

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer

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