The distance of the point (-1,9,-16) from the plane 2x+3y-z=5, measured parallel to the li…

Question

The distance of the point $(-1,9,-16)$ from the plane $2x+3y-z=5$, measured parallel to the line $\dfrac{x+4}{3}=\dfrac{2-y}{4}=\dfrac{z-3}{12}$, is:

Accepted Answer

Given, $${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$

$$ \Rightarrow $$ $${{x + 4} \over 3} = {{y - 2} \over -4} = {{z - 3} \over {12}}$$

Equation of line passing through the point P$(-1,9,-16)$ and parallel to line $\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}$ is $\frac{x+1}{3}=\frac{y-9}{-4}=\frac{z+16}{12}=\lambda$

Any point on this line (A) $=(3 \lambda-1,-4 \lambda+9,12 \lambda-16)$

Point of intersection line and plane

$$
\begin{aligned}
& 2(3 \lambda-1)+3(-4 \lambda+9)-1(12 \lambda-16)=5 \\
\Rightarrow & 6 \lambda-2-12 \lambda+27-12 \lambda+16=5 \\
\Rightarrow & \lambda=2 \quad \\
\therefore & \text { Point A }=(5,1,8) \\
\therefore & \text { Distance (AP) } \\
= & \sqrt{(5+1)^2+(9-1)^2+(-16-8)^2}\\
= & \sqrt{36+64+576}\\
= & 26 \text { units }
\end{aligned}
$$

Solutions · 1