Question

Find the distance of the point (1, –2, 3) from the plane x – y + z = 5, measured parallel to the line.

\(\frac{x-1}{2}=\frac{y-3}{3}=\frac{z+2}{-6}\)

Answer 1

Let Q \((\alpha, \beta, \gamma)\) be the point on the given plane

x - y + z = 5 .....(i)

Since PQ is parallel to given line

where P (1, -2, 3) is the given point.

\(\because\) PQ is parallel to given line (ii).

\(\therefore\) \(\vec{PQ}\big|\big| \vec{b}\) (parallel vector of line).

Now, \(\because\) Q \((\alpha, \beta, \gamma)\) lie on plane (i)

Therefore required distance