Question

The equation of the plane passing through the line of intersection of the planes $\overrightarrow{r}\cdot (2\hat{i}-3\hat{j}+4\hat{k})=1$ and $\overrightarrow{r}\cdot (\hat{i}-\hat{j})+4=0$ and perpendicular to the plane $\overrightarrow{r}\cdot (2\hat{i}-\hat{j}-\hat{k})+4=0,$ is

• A. $\overrightarrow{r}\cdot (2\hat{i}-\hat{j}+5\hat{k})=5$
• B. $\overrightarrow{r}\cdot (\hat{i}-2\hat{j}+4\hat{k})=3$
• C. $\overrightarrow{r}\cdot (\hat{i}-2\hat{j}+4\hat{k})=5$
• D. $\overrightarrow{r}\cdot (2\hat{i}-\hat{j}+5\hat{k})=3$

Answer 1

The correct option is C $\overrightarrow{r}\cdot (\hat{i}-2\hat{j}+4\hat{k})=5$

$\begin{array}{c}{\pi }_1=\overrightarrow{r}.\left(2\hat{i}-3\hat{j}+4\hat{k}\right)=1\Rightarrow 2x-3y+4z=1\\ {\pi }_2=\overrightarrow{r}.\left(\hat{i}-\hat{j}\right)+4=0\Rightarrow x-y+4=0\\ {\pi }_3=\overrightarrow{r}.\left(\hat{i}-\hat{j}-\hat{k}\right)+4=0\Rightarrow 2x-y-z+4=0\\ \left(2x-3y+4z-1\right)+\lambda \left(x-y+4\right)=0\\ Asitisperpendicularto2x-y-z+4=0\\ \Rightarrow 2\left(2+\lambda \right)-1\left(-3-\lambda \right)+4\times -1=0\\ 2\left(2+\lambda \right)+\left(3+\lambda \right)=4\\ 7+3\lambda =4\\ 3\lambda =-3\\ \Rightarrow \lambda =-1\\ equationofplane:x-2y+4z-5=0\\ \overrightarrow{r}\left(\hat{i}-2\hat{j}+4\hat{k}\right)=5\\ OptionCiscorrectanswer.\end{array}$