Question

The square of the distance of the image of the point $(6, 1, 5)$ in the line $\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4},$ from the origin is _____ .

Answer 1

Let $M(3\lambda + 1, 2\lambda, 4\lambda + 2)$

$\overrightarrow{AM} \cdot \mathbf{b} = 0$
$\implies 9\lambda - 15 + 4\lambda - 2 + 16\lambda - 12 = 0$
$\implies 29\lambda = 29 \implies \lambda = 1$

So,
$M(4, 2, 6), \quad I = (2, 3, 7)$

Hence Distance =
$\sqrt{4 + 9 + 49} = \sqrt{62}$

Therefore, the square of the distance of the image of the point is 62.