JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Three Dimensional Geometry

Let the foot of perpendicular from a point

P (1, 2, - 1)

to the straight line

L : \frac{x}{1} = \frac{y}{0} = \frac{z}{- 1}

N

. Let a line be drawn from

P

parallel to the plane

x + y + 2 z = 0

which meets

L

and point

Q

. If

α

is the acute angle between the lines PN and PQ , then

cos α

is equal to..... .

Answer: (C)

Solution

Points of

L

are

(r, 0, - r)

and its direction is

(1, 0, - 1)

. For the foot

N = (r, 0, - r)

, the vector

P N = (r - 1, - 2, 1 - r)

is perpendicular to

(1, 0, - 1)

(r - 1) + (r - 1) = 0 \Rightarrow r = 1

N = (1, 0, - 1)

and

P N = (0, - 2, 0)

.
The line

P Q

runs parallel to

x + y + 2 z = 0

, so

P Q

is perpendicular to the normal

(1, 1, 2)

. With

Q = (s, 0, - s)

L

P Q = (s - 1, - 2, 1 - s)

and

(s - 1) - 2 + 2 (1 - s) = 0 \Rightarrow - s - 1 = 0 \Rightarrow s = - 1

Q = (- 1, 0, 1)

and

P Q = (- 2, - 2, 2)

. Then

cos α = \frac{P N \cdot P Q}{P N P Q} = \frac{4}{2 \cdot 2 3}

and

cos α = \frac{1}{3}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option