Let P be the plane passing through the intersection of the planes r→·(i^+3j^-k^)=5 and r→·(2i^-j^+k^)=3, and the point (2,1,-2). Let the position vectors of the points X and Y be i^-2j^+4k^ and 5i^-j^+2k^ respectively. Then the points

Equation of plane passing through the line of intersection of planes P1=0,P2=0 is P1+λP2=0 ⇒(x+3y-z-5)+λ(2x-y+z-3)=0 (2,1,-2) lies on this plane ∴λ=1⇒plane is 3 x+2 y-8=0 sub the points X(1,-2,4) and Y(5,-1,2) in plane

Let P be the plane passing through the intersection of the planes r→·(i^+3j^-k^)=5 and r→·(2i^-j^+k^)=3, and the point (2,1,-2). Let the position vectors of the points X and Y be i^-2j^+4k^ and 5i^-j^+2k^ respectively. Then the points

Equation of plane passing through the line of intersection of planes P1=0,P2=0 is P1+λP2=0 ⇒(x+3y-z-5)+λ(2x-y+z-3)=0 (2,1,-2) lies on this plane ∴λ=1⇒plane is 3 x+2 y-8=0 sub the points X(1,-2,4) and Y(5,-1,2) in plane

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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

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X and X+Y are on the same side of P

X and Y are on the opposite sides of P

X + Y and X - Y are on the same side of P

Y and Y-X are on the opposite sides of P

answer is C.

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Detailed Solution

$E q u a t i o n o f p l a n e p a s \sin g t h r o u g h t h e l i n e o f i n t e r s e c t i o n o f p l a n e s P_{1} = 0, P_{2} = 0 i s P_{1} + λ P_{2} = 0 \Rightarrow (x + 3 y - z - 5) + λ (2 x - y + z - 3) = 0 (2, 1, - 2) l i e s o n t h i s p l a n e ∴ λ = 1 \Rightarrow p l a n e i s 3 x + 2 y - 8 = 0 s u b t h e p o i n t s X (1, - 2, 4) a n d Y (5, - 1, 2) i n p l a n e$