Question

Show that the line whose vector equation is $\overrightarrow{r}=2\hat{i}+5\hat{j}+7\hat{k}+\lambda \left(\hat{i}+3\hat{j}+4\hat{k}\right)$ is parallel to the plane whose vector equation is $\overrightarrow{r}·\left(\hat{i}+\hat{j}-\hat{k}\right)=7.$ Also, find the distance between them.

Answer 1

$$\begin{gathered} \mathrm{The\; given\; plane\; passes\; through\; the\; point}\text{with}position\; vector\overrightarrow{a}=2\hat{i}+5\hat{j}+7\hat{k}\text{and is parallel to the vector}\overrightarrow{b}=\hat{i}+3\hat{j}+4\hat{k}. \\ \text{The given plane is}\overrightarrow{r}\text{.}\left(\hat{i}+\hat{j}-\hat{k}\right)\text{= 7 or .}cc \\ \text{So, the normal vector,}\overrightarrow{n}\text{=}\hat{i}+\hat{j}-\hat{k}\text{and}d=7. \\ \text{Now,}\overrightarrow{b}.\overrightarrow{n}=\left(\hat{i}+3\hat{j}+4\hat{k}\right).\left(\hat{i}+\hat{j}-\hat{k}\right)=1+3-4=4-4=0 \\ \text{So,}\overrightarrow{b}\text{is perpendicular to}\overrightarrow{n}\text{.} \\ \text{So, the given line is parallel to the given plane.} \\ \text{The distance between the line and the parallel plane . Then,} \\ \text{d = length of the perpendicular from the point}\overrightarrow{a}=2\hat{i}+5\hat{j}+7\hat{k}\text{to}\mathrm{the}\mathrm{plane}\overrightarrow{\mathrm{r}}\text{.}\overrightarrow{\mathrm{n}}\text{=}\mathrm{d} \\ d=\frac{\left|\overrightarrow{a}.\overrightarrow{n}-d\right|}{\left|\overrightarrow{n}\right|} \\ =\frac{\left|\left(2\hat{i}+5\hat{j}+7\hat{k}\right).\left(\hat{i}+\hat{j}-\hat{k}\right)-7\right|}{\left|\hat{i}+\hat{j}-\hat{k}\right|} \\ =\frac{\left|2+5-7-7\right|}{\sqrt{1+1+1}} \\ =\frac{7}{\sqrt{3}}\text{units} \end{gathered}$$