Question

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

• A. $-2\hat{i}+7\hat{j}+13\hat{k}$
• B. $2\hat{i}+7\hat{j}-13\hat{k}$
• C. $-2\hat{i}-7\hat{j}+13\hat{k}$
• D. $2\hat{i}+7\hat{j}+13\hat{k}$

Answer 1

The correct option is A $-2\hat{i}+7\hat{j}+13\hat{k}$

Given:

$P_1=r\cdot (3\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})=1$

$P_2=r\cdot (\overrightarrow{i}+4\overrightarrow{j}-2\overrightarrow{k})=2$

We know that :

The parallel vector of the line of intersection of planes is:

$\overrightarrow{b}=n_1\times n_2$

and

$a\times b=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$

Solution:

here,

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

and

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

Now ,

$\Rightarrow \overrightarrow{b}=n_1\times n_2$

$\Rightarrow \overrightarrow{b}=(3\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})\times (\overrightarrow{i}+4\overrightarrow{j}-2\overrightarrow{k})$

$\Rightarrow \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 3& -1& 1\\ 1& 4& -2\end{array}\right|$

$\Rightarrow \overrightarrow{b}=\left|\begin{array}{cc}-1& 1\\ 4& -2\end{array}\right|\overrightarrow{i}-\left|\begin{array}{cc}3& 1\\ 1& -2\end{array}\right|\overrightarrow{j}+\left|\begin{array}{cc}3& -1\\ 1& 4\end{array}\right|\overrightarrow{k}$

$\Rightarrow \overrightarrow{b}=\left[\right(-1)\cdot (-2)-1\cdot 4]\overrightarrow{i}-[3\cdot (-2)-1\cdot 1]\overrightarrow{j}+[3\cdot 4-1\cdot (-1\left)\right]\overrightarrow{k}$

$\Rightarrow \overrightarrow{b}=[2-4]\overrightarrow{i}-\left[\right(-6)-1]\overrightarrow{j}+[12-(-1\left)\right]\overrightarrow{k}$

$\Rightarrow \overrightarrow{b}=(-2)\overrightarrow{i}-(-7)\overrightarrow{j}+13\overrightarrow{k}$

$\Rightarrow \overrightarrow{b}=-2\overrightarrow{i}+7\overrightarrow{j}+13\overrightarrow{k}$