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) $= 2 i +\hat{ k }$ 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
$r \cdot( b \times c )= a \cdot( b \times c )=[ a b c ]$
Now,
$\left[a b c\right]=\begin{vmatrix}2&0&1\\ 1&0&0\\ 1&2&-3\end{vmatrix}$
$=2\left(0\right)-0+1\left(2-0\right)=2$
and $b\times c =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 1&0&0\\ 1&2&-3\end{vmatrix} =3\hat{i} +2\hat{k}$
$\therefore r \cdot(3 \hat{i}+2 \hat{ k })=2$
Therefore, $\alpha=2$