Consider the equation of line AB is

MATHEMATICS · Single correct · CBSE · Mathematics · Three-Dimensional Geometry · Three-Dimensional Geometry

Problem

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

. Through a point

P(1, 2, 5)

line PN is drawn perpendicular to AB and line PQ is drawn parallel to the plane

3x + 4y + 5z = 0

to meet AB is Q. Then, (1) coordinate of

N

are

\left( \frac{52}{49}, -\frac{78}{49}, \frac{156}{49} \right)

(2) the coordinates of

Q

are

\left( 3, -\frac{9}{2}, 9 \right)

(3) the equation of

PN

\frac{x-1}{3} = \frac{y-2}{-176} = \frac{z-5}{-89}

(4) coordinates of

N

are

\left( \frac{156}{49}, \frac{52}{49}, -\frac{78}{49} \right)