Question

The vector equation of the plane through the point (2, 1, –1) and passing through the line of intersection of the plane $\overline{r}.(\hat{i}+3\hat{j}-\hat{k})=0$ and $\overline{r}=(\hat{j}+2\hat{k})=0$ , is :
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• A. $\overline{r}.(\hat{i}+9\hat{j}+11\hat{k})=0$
• B. $\overline{r}.(\hat{j}-9\hat{j}-11\hat{k})=6$
• C. $\overline{r}.(\hat{i}-3\hat{j}-13\hat{k})=0$
• D. $\overline{r}.(\hat{i}-3\hat{j}-8\hat{k})=0$

Answer 1

The correct option is A $\overline{r}.(\hat{i}+9\hat{j}+11\hat{k})=0$

Equation of plane passing through the line of intersection of given planes is
$\overline{r}.(\hat{i}+3\hat{j}-\hat{k})+\lambda (\overline{r}.(\hat{j}+2\hat{k}\left)\right)=0$
This plane passes through point (2, 1, – 1). Therefore
$(2\hat{i}+\hat{j}-\hat{k}).(\hat{i}+3\hat{j}-\hat{k})+\lambda (2\hat{i}+\hat{j}-\hat{k}).(\hat{j}+2\hat{k})=0$
$\Rightarrow 6-\lambda =0$ i.e., $\lambda =6$.
Hence, equation of required plane is
$\overline{r}.(\hat{i}+9\hat{j}+11\hat{k})=0$