Question

Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured parallel to the line whose direction cosines are proportional to 2, 3, −6.

Answer 1

The equation of the given plane is x − y + z = 5.
We have to find the distance of point (1, −2, 3) from the plane measured parallel to line whose direction cosines are proportional to 2, 3, −6.

 

Let this line meet the plane at Q.

The equation of the line passing through (1,−2, 3) and having direction cosines proportional to 2, 3, −6 is

Let this line meet the plane at Q.
The equation of the line passing through (1,−2, 3) and having direction cosines proportional to 2, 3, −6 is \[\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} = \lambda\]

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

So, the coordinates of any point on the line are in the form \[\left( 2\lambda + 1, 3\lambda - 2, - 6\lambda + 3 \right)\]

If this point \[\left( 2\lambda + 1, 3\lambda - 2, - 6\lambda + 3 \right)\] lies on the plane x − y + z = 5, then \[2\lambda + 1 - 3\lambda + 2 - 6\lambda + 3 = 5\]

\[\Rightarrow - 7\lambda + 6 = 5\]

\[\Rightarrow - 7\lambda = - 1\]

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

Therefore, the coordinates of point Q are

\[\left( \frac{2}{7} + 1, \frac{3}{7} - 2, - \frac{6}{7} + 3 \right)\] i.e. 

\[\left( \frac{9}{7}, - \frac{11}{7}, \frac{15}{7} \right)\] .

Using distance formula, we have

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

Thus, the required distance is 1 unit.