Question

Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to  \[\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .\]

 

Answer 1

The given plane is x − y + z = 5. We need to find the distance of the point from the point(1, -2, 3) measured along a parallel line 

\[\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .\]

Let the line from point be P(1, -2, 3) and meet the plane at point Q.
Direction ratios of the line from the point ​(1, -2, 3) to the given plane will be the same as the given line

\[\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .\]

So the equation of the line passing through P and with same direction ratios will be:

\[\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} = \lambda\]

\[ \Rightarrow x = 2\lambda + 1, y = 3\lambda - 2, z = - 6\lambda + 3\]

coordinates of any point on the line PQ are 

\[x = 2\lambda + 1, y = 3\lambda - 2, z = - 6\lambda + 3\]

Now, since Q lies on the plane so it must satisfy the equation of the plane.

that is, ​x − y + z = 5

therefore, 2λ+1 - (3λ - 2) + (-6λ +3) = 5

 

\[\lambda = \frac{1}{7}\]

coordinates of Q are 

\[(\frac{2}{7} + 1), (\frac{3}{7} - 2), (\frac{- 6}{7} + 3)\]

\[\frac{9}{7}, \frac{- 11}{7}, \frac{15}{7}\]

using distance formula we have the length of PQ as

\[PQ = \sqrt{\left( \frac{9}{7} - 1 \right)^2 + \left( \frac{- 11}{7} + 2 \right)^2 + \left( \frac{15}{7} - 3 \right)^2}\]
\[ = \sqrt{\left( \frac{2}{7} \right)^2 + \left( \frac{3}{7} \right)^2 + \left( \frac{- 6}{7} \right)^2}\]
\[ = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}}\]
\[ = \sqrt{\frac{49}{49}} = 1\]

Hence PQ = 1
So, ​the distance of the point (1, −2, 3) from the plane x − y + z = 5  is 1