Question

Find the distance of the point (1, $-$5, 9) from the plane $x-y+z=$5 measured along the line $x=y=z$.

Answer 1

The equation of line parallel to the line x = y = z and passing through the point (1, −5, 9) is

$\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}$ .....(1)

Any point on this line is of the form (k + 1, k − 5, k + 9).

If (k + 1, k − 5, k + 9) be the point of intersection of line (1) and the given plane, then

(k + 1) − (k − 5) + (k + 9) = 5

⇒ k = −10

So, the point of intersection of line (1) and the given plane is (−10 + 1, −10 − 5, −10 + 9) i.e. (−9, −15, −1).

∴ Required distance = Distance between (1, −5, 9) and (−9, −15, −1) = $\sqrt{{\left(1+9\right)}^2+{\left(-5+15\right)}^2+{\left(9+1\right)}^2}=\sqrt{3\times {10}^2}=10\sqrt{3}$ units