Question

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line $\overrightarrow{r}=\hat{i}+3\hat{j}-2\hat{k}+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right).$

Answer 1

$$\begin{gathered} \text{The equation of any plane passing through (3, 4, 2) is} \\ a\left(x-3\right)+b\left(y-4\right)+c\left(z-2\right)=0...\left(1\right) \\ \text{It is given that (1) is passing through (7, 0, 6). So,} \\ a\left(7-3\right)+b\left(0-4\right)+c\left(6-2\right)=0 \\ \Rightarrow 4a-4b+4c=0 \\ \Rightarrow a-b+c=0...\left(2\right) \\ \text{It is given that (1) is perpendicular to the plane 2}x-5y+0z+15z=0.\text{So,} \\ 2a-5b+0c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x-3& y-4& z-2\\ 1& -1& 1\\ 2& -5& 0\end{array}\right|=0 \\ \Rightarrow 5\left(x-3\right)+2\left(y-4\right)-3\left(z-2\right)=0 \\ \Rightarrow 5x+2y-3z=17 \\ \text{Or}\overrightarrow{r}\text{.}\left(5\hat{i}+2\hat{j}-3\hat{k}\right)\text{= 17} \\ \mathbf{\text{Showing that the plane contains the line}} \\ \text{The line}\overrightarrow{r}=\left(\hat{i}+3\hat{j}-2\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)\text{passes through a point whose positon vector is}\overrightarrow{a}\text{=}\hat{i}+3\hat{j}-2\hat{k}\text{and is parallel to the vector}\overrightarrow{b}\text{=}\hat{i}-\hat{j}+\hat{k}. \\ \text{If the plane}\overrightarrow{r}\text{.}\left(5\hat{i}+2\hat{j}-3\hat{k}\right)\text{=17 contains the given line, then} \\ \text{(1) it should pass through the point}\hat{i}+3\hat{j}-2\hat{k} \\ \text{(2) it should be parallel to the line} \\ \text{Now,}\left(\hat{i}+3\hat{j}-2\hat{k}\right)\text{.}\left(5\hat{i}+2\hat{j}-3\hat{k}\right)\text{= 5 + 6 + 6 = 17} \\ \text{So, the plane passes through the point}\hat{i}+3\hat{j}-2\hat{k}. \\ \text{The normal vector to the given plane is}\overrightarrow{n}\text{=}\hat{i}-\hat{j}+\widehat{k.} \\ \text{We observe that} \\ \overrightarrow{b}.\overrightarrow{n}=\left(\hat{i}-\hat{j}+\hat{k}\right).\left(5\hat{i}+2\hat{j}-3\hat{k}\right)=5-2-3=0 \\ \text{Therefore, the plane is parallel to the line.} \\ \text{Hence, the given plane contains the given line.} \end{gathered}$$