Question

Let $P$ be the plane passing through the intersection of the planes $\vec{r}\cdot (\hat{i}+3\hat{j}-\hat{k})=5$ and $\vec{r}\cdot (2\hat{i}-\hat{j}+\hat{k})=3$, and the point $(2,1,-2)$. Let the position vectors of the points $X$ and $Y$ be $\hat{i}-2\hat{j}+4\hat{k}$ and $5\hat{i}-\hat{j}+2\hat{k}$ respectively. Then the points

• A. $X$ and $X+Y$ are on the same side of $P$
• B. $Y$ and $Y-X$ are on the opposite sides of $P$
• C. $X$ and $Y$ are on the opposite sides of $P$
• D. $X+Y$ and $X-Y$ are on the same side of $P$

Answer 1

Answer: (C)

$P_1+\lambda P_2=0$
$\Rightarrow (x+3y-z-5)+\lambda (2x-y+z-3)=0$
$(2,1,-2)$ lies on this plane
$\therefore \lambda =1$
$\Rightarrow$ plane is $3x+2y-8=0$