Question

The line of intersection of the planes $\overrightarrow{r}.(3i-j+k)=1$ and $\overrightarrow{r}.(i+4j-2k)=2$ is parallel to the vector

• A. $2i+7j+13k$
• B. $-2i-7j+13k$
• C. $2i+7j-13k$
• D. $-2i+7j+13k$

Answer 1

The correct option is D $-2i+7j+13k$

the eq of line of intersection is

$\overrightarrow{r}=(x\hat{i}+y\hat{j}+z\hat{k})+\lambda (\overrightarrow{n_1}\times \overrightarrow{n_2})$

where $\overrightarrow{n_1}and\overrightarrow{n_2}$ are normal vector of plane $P_{1}and{P}_2$

line of intersection of planes is parallel to $\overrightarrow{n_1}\times \overrightarrow{n_2}$

$\overrightarrow{n_1}=3\hat{i}-\hat{j}+\hat{k}$

$\overrightarrow{n_2}=\hat{i}+4\hat{j}-2\hat{k}$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ 1& 4& -2\end{array}\right|$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\hat{i}(2-4)-\hat{j}(-6-1)+\hat{k}(12+1)$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=-2\hat{i}+7\hat{j}+13\hat{k}$