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:
$$\begin{gathered} \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=\lambda \\ \Rightarrow x=2\lambda +1,y=3\lambda -2,z=-6\lambda +3 \end{gathered}$$
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
$$\begin{gathered} PQ=\sqrt{{\left(\frac{9}{7}-1\right)}^2+{\left(\frac{-11}{7}+2\right)}^2+{\left(\frac{15}{7}-3\right)}^2} \\ =\sqrt{{\left(\frac{2}{7}\right)}^2+{\left(\frac{3}{7}\right)}^2+{\left(\frac{-6}{7}\right)}^2} \\ =\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}} \\ =\sqrt{\frac{49}{49}}=1 \end{gathered}$$

Hence PQ = 1

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