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