Question

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

• A. $r.(2\hat{i}+7\hat{j}-13\hat{k})=1$
• B. $r.(2\hat{i}-7\hat{j}-13\hat{k})=1$
• C. $r.(2\hat{i}+7\hat{j}+13\hat{k})=0$
• D. None of the above

Answer 1

The correct option is B $r.(2\hat{i}-7\hat{j}-13\hat{k})=1$

The line of intersection of the planes
$r.(3\hat{i}-\hat{j}+\hat{k})=1;r.(\hat{i}+4\hat{j}-2\hat{k})=2$
is common to both the parties
Therefore, it is perpendicular to normals to the planes $n_1=3\hat{i}-\hat{j}+\hat{k};n_2=\hat{i}+4\hat{j}-2\hat{k}$
Hence it is parallel to the vector
$n_1\times n_2=-2\hat{i}+7\hat{j}+13\hat{k}$
Thus, we have to find the equation of the plane passing through $a=\hat{i}+2\hat{j}-\hat{k}$ and normal to the vector $n=n_1\times n_2$
The equation of the required plane is
$(r-a).n\Rightarrow r.n=a.n$
$r.(-2\hat{i}+7\hat{j}+13\hat{k})=(\hat{i}+2\hat{j}-\hat{k}).(-2\hat{i}+7\hat{j}+13\hat{k})$
$r.(-\hat{i}-7\hat{j}-13\hat{k})=1$