JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Three Dimensional Geometry

The foot of the perpendicular from a point on the circle

x^{2} + y^{2} = 1, z = 0

to the plane

2 x + 3 y + z = 6

lies on which one of the following curves?

(A) $(6 x + 5 y - 12)^{2} + 4 (3 x + 7 y - 8)^{2} = 1$ , $z = 6 - 2 x - 3 y$
(B) $(5 x + 6 y - 12)^{2} + 4 (3 x + 5 y - 9)^{2} = 1$ , $z = 6 - 2 x - 3 y$ ✓ Correct
(C) $(6 x + 5 y - 14)^{2} + 9 (3 x + 5 y - 7)^{2} = 1$ , $z = 6 - 2 x - 3 y$
(D) $(5 x + 6 y - 14)^{2} + 9 (3 x + 7 y - 8)^{2} = 1$ , $z = 6 - 2 x - 3 y$

Answer: (B)

Solution

A point on the circle is

(cos θ, sin θ, 0)

. The perpendicular from this point to the plane

2 x + 3 y + z = 6

has direction ratios

(2, 3, 1)

.

The foot of perpendicular

(h, k, w)

satisfies:

\frac{h - c o s θ}{2} = \frac{k - s i n θ}{3} = \frac{w - 0}{1} = \frac{- ( 2 c o s θ + 3 s i n θ - 6 )}{14}

Solving:

h = \frac{10 c o s θ - 6 s i n θ + 12}{14}

k = \frac{5 s i n θ - 6 c o s θ + 18}{14}

.

Eliminating

sin θ

and

cos θ

using

sin^{2} θ + cos^{2} θ = 1

gives:

(5 h + 6 k - 12)^{2} + 4 (3 h + 5 k - 9)^{2} = 1

with

z = 6 - 2 x - 3 y

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option