Question

The vector equation of the plane
$r=\left(2\hat{i}+\hat{k}\right)+\lambda \left(\hat{i}\right)+\mu \left(\hat{i}+2\hat{j}-3\hat{k}\right)$ in scalar product form is $r\cdot \left(3\hat{i}+2\hat{k}\right)=\alpha$, then $\alpha =...$

• A. $2$
• B. $3$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

Given, equation of plane
$r=(2\hat{i}+\hat{k})+\lambda \hat{i}+\mu (\hat{i}+2\hat{j}-3\hat{k})$...(i)
Here, plane (i) passing through a (let) $=2i+\hat{k}$ and parallel to vector $b($ let $)=\hat{i}$ and
$c=\hat{i}+2\hat{j}-3\hat{k}$
We know that equation of plane passing through a point a and parallel to non-parallel vectors $b$ and $c$ is
$r\cdot (b\times c)=a\cdot (b\times c)=[abc]$
Now,
$\left[abc\right]=\left|\begin{array}{ccc}2& 0& 1\\ 1& 0& 0\\ 1& 2& -3\end{array}\right|$
$=2\left(0\right)-0+1\left(2-0\right)=2$
and $b\times c=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& 0\\ 1& 2& -3\end{array}\right|=3\hat{i}+2\hat{k}$
$\therefore r\cdot (3\hat{i}+2\hat{k})=2$
Therefore, $\alpha =2$