Question

The vector equation of the plane through the point $\hat{i}+2\hat{j}-\hat{k}$ & perpendicular to the line of intersection of the planes $\overline{r}.(3\hat{i}-\hat{j}+\hat{k})=1$ and $\overline{r}.(\hat{i}+4\hat{j}-2\hat{k})=2$ is

• A. $\overline{r}.(2\hat{i}+7\hat{j}-13\hat{k})=29$
• B. $\overline{r}.(2\hat{i}-7\hat{j}-13\hat{k})=1$
• C. $\overline{r}.(2\hat{i}-7\hat{j}-+3\hat{k})=25$
• D. $\overline{r}.(2\hat{i}+7\hat{j}-+3\hat{k})=3$

Answer 1

Answer: (B)

$\overline{r}.(2\hat{i}-7\hat{j}-13\hat{k})=1$

The line of intersection of planes $\overline{r}\cdot (3\hat{i}-\hat{j}+\hat{k})=1$ and $\overline{r}\cdot (\hat{i}+4\hat{j}-2\hat{k})=2$ is common to both the planes, so line of intersection is $\perp$er to the nonnal to the two planes
i.e. ${\overline{n}}_1=3\hat{i}-\hat{j}+\hat{k}and{\overline{n}}_2=\hat{i}+4\hat{j}-2\hat{k}$
Hence it is parallel to the vector ${\overline{n}}_1\times {\overline{n}}_2$
where ${\overline{n}}_1\times {\overline{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ 1& 4& -2\end{array}\right|=-2\hat{i}+7\hat{j}+13\hat{k}$

Now required plane $(\overline{r}-\overline{a})\cdot \overline{n}=0;\overrightarrow{a}=\hat{i}+2\hat{j}-\hat{k};\overrightarrow{n}=-2\hat{i}+7\hat{j}+13\hat{k}$
$\Rightarrow \overline{r}\cdot \overline{n}=\overline{a}\cdot \overline{n}$
$\Rightarrow \overline{r}\cdot (-2\hat{i}+7\hat{j}+13\hat{k})=-1$
$\Rightarrow \overline{r}\cdot (2\hat{i}-7\hat{j}-13\hat{k})=1$
Hence (B) is correct choice.