Let A and B two distinct points on the line L: x-6/3 = y-7/3 = z-7/-2.

Question

Let A and B two distinct points on the line \(L: \frac{x-6}{3} = \frac{y-7}{3} = \frac{z-7}{-2}.\) Both A and B are at a distance \(2\sqrt{22}\) from the foot of the perpendicular drawn from the point (1, 2, 3) on the line L. If O is origin then \(\overrightarrow{OA}. \overrightarrow{OB}\) is equal to

Answer 1

Answer is: 18 

\( \overrightarrow {PM} = 3 \lambda + 5, 3\lambda+5, -2\lambda +4\) 

\(\overrightarrow L = (3, 3, -2)\) 

\(\overrightarrow {PM}. \overrightarrow L = 0 = 9\lambda +15+9\lambda + 15 + 4\lambda - 8 = 0\)  

\(22 \lambda + 22 = 0 \Rightarrow \lambda = -1\)

M (3, 4, 9)

Now,