Let A and B be two distinct points on the line \(\mathrm{L}: \frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-7}{-2}\). Both A and B are at a distance \(2 \sqrt{17}\) from the foot of perpendicular drawn from the point \((1,2,3)\) on the line L. If O is the origin, then \(\overrightarrow{OA} \cdot \overrightarrow{OB}\) is equal to:

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Three Dimensional Geometry

Let A and B be two distinct points on the line

L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}

. Both A and B are at a distance

217

from the foot of perpendicular drawn from the point

(1, 2, 3)

on the line L. If O is the origin, then

O A \cdot O B

is equal to:

Answer: (B)

The line passes through

(6, 7, 7)

with direction

(3, 2, - 2)

, so a general point is

(6 + 3 t, 7 + 2 t, 7 - 2 t)

. The foot

Q

of the perpendicular from

(1, 2, 3)

satisfies

(5 + 3 t, 5 + 2 t, 4 - 2 t) \cdot (3, 2, - 2) = 0

, that is

17 t + 17 = 0

, so

t = - 1

and

Q = (3, 5, 9)

. For a point at parameter

s

the squared distance from

Q

17 (s + 1)^{2}

; setting it equal to

(217)^{2} = 68

gives

(s + 1)^{2} = 4

, so

s = 1

s = - 3

. These give

A = (9, 9, 5)

and

B = (- 3, 1, 13)

, and

O A \cdot O B = - 27 + 9 + 65 = 47

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option