Question

Show that the plane whose vector equation is $\overrightarrow{r}·\left(\hat{i}+2\hat{j}-\hat{k}\right)=1$ and the line whose vector equation is $\overrightarrow{r}=\left(-\hat{i}+\hat{j}+\hat{k}\right)+\lambda \left(2\hat{i}+\hat{j}+4\hat{k}\right)$ are parallel. Also, find the distance between them.

Answer 1

$$\begin{gathered} \mathrm{The\; given\; plane\; passes\; through\; the\; point}\text{with}position\; vector\overrightarrow{a}=-\hat{i}+\hat{j}+\hat{k}\text{and is parallel to the vector}\overrightarrow{b}=2\hat{i}+\hat{j}+4\hat{k}. \\ \text{The given plane is}\overrightarrow{r}\text{.}\left(\hat{i}+2\hat{j}-\hat{k}\right)\text{= 1 or}\overrightarrow{r}\text{.}\overrightarrow{n}\text{=}d. \\ \text{So, normal vector,}\overrightarrow{n}\text{=}\hat{i}+2\hat{j}-\hat{k}\text{and}d=1 \\ \text{Now,}\overrightarrow{b}.\overrightarrow{n}=\left(2\hat{i}+\hat{j}+4\hat{k}\right).\left(\hat{i}+2\hat{j}-\hat{k}\right)=2+2-4=0 \\ \text{So},\overrightarrow{b}\text{is perpendicular to}\overrightarrow{n}\text{.} \\ \text{So, the given line is parallel to the given plane.} \\ \text{The distance between the line and the parallel plane is the distance between any point on the line and the given plane. Since the plane passes through the point}\overrightarrow{a}=-\hat{i}+\hat{j}+\widehat{k,} \\ \text{the perpendicular distance from the given line to the plane is} \\ d=\frac{\left|\overrightarrow{a}.\overrightarrow{n}-d\right|}{\left|\overrightarrow{n}\right|} \\ =\frac{\left|\left(-\hat{i}+\hat{j}+\hat{k}\right).\left(\hat{i}+2\hat{j}-\hat{k}\right)-1\right|}{\left|\hat{i}+2\hat{j}-\hat{k}\right|} \\ =\frac{\left|-1+2-1-1\right|}{\sqrt{1+4+1}} \\ =\frac{1}{\sqrt{6}}\text{units} \end{gathered}$$