The distance of the point $(1,-5,9)$ from the plane $x-y+z=5$ measured along the line $x=y=z$ is ______ units.
The distance of the point $(1,-5,9)$ from the plane $x-y+z=5$ measured along the line $x=y=z$ is ______ units.
$3 \sqrt{10}$
$10 \sqrt{3}$
$\frac{10}{\sqrt{3}}$
$\frac{20}{3}$
Solution
Equation of the line can be written as $\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda$
$\therefore \quad$ Co-ordinates of any point on the line are given as $x=\lambda+1, y=\lambda-5, \mathrm{z}=\lambda+9$
Substituting in the equation of the plane, we get
$\begin{array}{ll}
& (\lambda+1)-(\lambda-5)+(\lambda+9)=5 \\
\therefore & \lambda+15=5 \\
\therefore \quad & \lambda=-10 \\
\therefore \quad & \text { Point on the plane is }(-9,-15,-1) \\
\therefore \quad & \text { Required distance } \\
& =\text { Distance between }(1,-5,9) \text { and }(-9,-15,-1) \\
& =\sqrt{(-9-1)^2+(-15+5)^2+(-1-9)^2} \\
& =10 \sqrt{3}
\end{array}$