JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P\) be the plane passing through the intersection of the planes
\(\overrightarrow{ r } \cdot(\hat{ i }+3 \hat{ j }-\hat{ k })=5\) and \(\overrightarrow{ r } \cdot(2 \hat{ i }-\hat{ j }+\hat{ k })=3\), 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 & Solution
Correct Answer
(C) \(X\) and \(Y\) are on the opposite sides of \(P\)
Step-by-step Solution
Detailed explanation
\(P _{1}+\lambda P _{2}=0\) \(\Rightarrow( x +3 y - z -5)+\lambda(2 x - y + z -3)=0\) \((2,1,-2)\) lies on this plane \(\therefore \lambda=1 \Rightarrow\) plane is \(3 x +2 y -8=0\)
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