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
$\begin{vmatrix} x-0 & y-0 & z-0 \\ 0-0 & 0-0 & 4-0 \\ 2-0 & 1-0 & 0-0 \end{vmatrix}=0$
$\Rightarrow \begin{vmatrix} x & y & z \\ 0 & 0 & 4 \\ 2 & 1 & 0 \end{vmatrix}=0$
$\Rightarrow -4(x-2 y)=0 \Rightarrow x-2 y=0$
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. $(-8,-4,7)$
$\therefore O A =-8 \hat{ i }-4 \hat{ j }+7 \hat{ k }$