JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Three Dimensional Geometry

A perpendicular is drawn from a point on the line

\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z}{1}

to the plane

x + y + z = 3

such that the foot of the perpendicular

Q

also lies on the plane

x - y + z = 3

. Then the co-ordinates of Q are

(A) $(2, 0, 1)$ ✓ Correct
(B) $(- 1, 0, 4)$
(C) $(1, 0, 2)$
(D) $(4, 0, - 1)$

Answer: (A)

Solution

A point on the line is

P (2 λ + 1, - λ - 1, λ)

, and its foot of perpendicular

Q

x + y + z = 3

is required also to satisfy

x - y + z = 3

. Subtracting the two plane equations forces

y = 0

Q

. Carrying the perpendicular, whose direction is

(1, 1, 1)

, down from

P

and imposing

y = 0

together with

x + z = 3

pins the parameter at

λ = 0

, which places

Q

(2, 0, 1)

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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