Question

A line $L$ passes through the points $\hat{i}+2\hat{j}+\hat{k}$ and $-2\hat{i}+3\hat{k}$. A plane $P$ passes through the origin and the points $4\hat{k},2\hat{i}+\hat{j}$. The point where the line $L$ meets the plane $P$ is

• A. $-\hat{i}-\hat{j}+3\hat{k}$
• B. $-8\hat{i}-4\hat{j}+7\hat{k}$
• C. $8\hat{i}+4\hat{j}+\hat{k}$
• D. $3\hat{i}+\hat{j}+2\hat{k}$

Answer 1

Answer: (B)

Equation of line passing through $(1,2,1)$ and $(-2,0,3)$ is given by
$\frac{x-1}{-2-1}=\frac{y-2}{0-2}=\frac{z-1}{3-1}=\lambda \text{ (say) }$
$\Rightarrow \frac{x-1}{-3}=\frac{y-2}{-2}=\frac{z-1}{2}=\lambda$
Any point on this line has coordinate $A(-3\lambda +1,-2\lambda +22\lambda +1)$
Equation of plane passing through the points $(0,0,0),(0,0,4)$ and $(2,1,0)$ is given by
$\left|\begin{array}{ccc}x-0& y-0& z-0\\ 0-0& 0-0& 4-0\\ 2-0& 1-0& 0-0\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x& y& z\\ 0& 0& 4\\ 2& 1& 0\end{array}\right|=0$
$\Rightarrow -4(x-2y)=0\Rightarrow x-2y=0$
Since, $A$ lies on above plane
$\therefore -3\lambda +1-2(-2\lambda +2)=0$
$\Rightarrow -3\lambda +1+4\lambda -4=0$
$\Rightarrow \lambda -3=0\Rightarrow \lambda =3$
$\therefore$ Coordinates of point $A$ are
$(-3\times 3+1,-2\times 3+2$ $2\times 3+1)$, i.e. $(-8,-4,7)$
$\therefore OA=-8\hat{i}-4\hat{j}+7\hat{k}$