Question

The distance of the point $(1,0,-3)$ from the plane $x-y$ $-z=9$ measured parallel to the line;
$\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$ is

• A. 10 units
• B. 9 units
• C. 8 units
• D. 7 units

Answer 1

Answer: (D)

The given plane is $x-y-z=9$ .... (i)
The given line $AB$ is $\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$ .... (ii)

The equation of the line passing through $(1,0,-3)$ and parallel to
$\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$ is
$\frac{x-1}{2}=\frac{y-0}{3}=\frac{z+3}{-6}=r$ .... (iii)
Coordinate of any point on (iii) may be given as
$P(2r+1,3r,-6r-3)$.
If $P$ is the point of the intersection of (i) and (iii), then it must lie on (i). Therefore,
$(2r+1)-(3r)-(-6r-3)=9$
$2r+1-3r+6r+3=9$ or $r=1$
Therefore, the coordinates of $P$ are $3,3,-9$.
Distance between $Q(1,0,-3)$ and $P(3,3,-9)$
$=\sqrt{(3-1{)}^2+(3-0{)}^2+(-9+3{)}^2}$
$=\sqrt{4+9+36}=7$