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. $6$
• B. $7$
• C. $8$
• D. $9$

Answer 1

Answer: (B)

$7$

Now, the equation of the line passing through $(\mathrm{1.0.}-3)$ and parallel to the line $\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$ is $\frac{x-1}{2}=\frac{y-0}{3}=\frac{z+3}{-6}$.......(1) which will intersect the plane $x-y-z=9$........(2).

To find the point of intersection of line (1) and plane (2), let us consider any point on the line be $(2r+1,3r,-6r-3)$. This point lies on the plane (2).

Then

$(2r+1)-\left(3r\right)-(-6r-3)=9$

or, $5r=5$

or, $r=1$.

The point of intersection is $(3,3,-9)$.

Now, the distance between the points $(1,0,-3)$ and $(3,3,-9)$ $=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7$, which is the required distance