Question

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

• A. $\overrightarrow{r}\cdot (\hat{i}-7\hat{j}+3\hat{k})=\frac{7}{3}$
• B. $\overrightarrow{r}\cdot (\hat{i}+7\hat{j}+3\hat{k})=7$
• C. $\overrightarrow{r}\cdot (3\hat{i}+7\hat{j}+3\hat{k})=7$
• D. $\overrightarrow{r}\cdot (\hat{i}+7\hat{j}+3\hat{k})=\frac{7}{3}$

Answer 1

The correct option is B $\overrightarrow{r}\cdot (\hat{i}+7\hat{j}+3\hat{k})=7$

Family of planes passing through intersection of the given planes is
$\{\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})-1\}+\lambda \{\overrightarrow{r}\cdot (\hat{i}-2\hat{j})+2\}=0$
The above curve passes through $\hat{i}+2\hat{k}.$
$(3-1)+\lambda (1+2)=0$
$\Rightarrow \lambda =-\frac{2}{3}$
Hence, equation of the plane is
$3\{\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})-1\}-2\{\overrightarrow{r}\cdot (\hat{i}-2\hat{j})+2\}=0$
$\Rightarrow \overrightarrow{r}\cdot (\hat{i}+7\hat{j}+3\hat{k})=7$

$TRICK:$ Only option $\overrightarrow{r}\cdot (\hat{i}+7\hat{j}+3\hat{k})=7$ satisfies the point $(1,0,2)$