Question

Show that the plane whose vector equation is $\overrightarrow{r}·\left(\hat{i}+2\hat{j}-\hat{k}\right)=3$ contains the line whose vector equation is $\overrightarrow{r}=\hat{i}+\hat{j}+\lambda \left(2\hat{i}+\hat{j}+4\hat{k}\right).$

Answer 1

$$\begin{gathered} \text{The line}\overrightarrow{r}=\left(\hat{i}+\hat{j}+0\hat{k}\right)+\lambda \left(2\hat{i}+\hat{j}+4\hat{k}\right)....\left(1\right)\text{passes through a point whose position vector is}\overrightarrow{a}\text{=}\hat{i}+\hat{j}+0\hat{k}\text{and is parallel to the vector}\overrightarrow{b}\text{=}2\hat{i}+\hat{j}+4\hat{k}. \\ \text{If the plane}\overrightarrow{r}\text{.}\left(\hat{i}+2\hat{j}-\hat{k}\right)\text{=3 contains the given line, then} \\ \text{(1) it should passes through the point}\hat{i}+\hat{j}+0\hat{k} \\ \text{(2) it should be parallel to the line} \\ \text{Now,}\left(\hat{i}+\hat{j}+0\hat{k}\right)\text{.}\left(\hat{i}+2\hat{j}-\hat{k}\right)\text{= 1 + 2 = 3} \\ \text{So, the plane passes through the point}\hat{i}+\hat{j}+0\hat{k}. \\ \text{The normal vector to the given plane is}\overrightarrow{n}\text{=}\hat{i}+2\hat{j}-\widehat{k.} \\ \text{We observe that} \\ \overrightarrow{b}.\overrightarrow{n}=\left(2\hat{i}+\hat{j}+4\hat{k}\right).\left(\hat{i}+2\hat{j}-\hat{k}\right)=2+2-4=0 \\ \text{Therefore, the plane is parallel to the line.} \\ \text{Hence, the given plane contains the given line.} \end{gathered}$$