JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the plane \(P: 4 x-y+z=10\) be rotated by an angle \(\frac{\pi}{2}\) about its line of intersection with the plane \(x+y-z=4\). If \(\alpha\) is the distance of the point \((2,3,-4)\) from the new position of the plane \(P\), then \(35 \alpha\) is
- A \(90\)
- B \(85\)
- C \(105\)
- D \(126\)
Answer & Solution
Correct Answer
(D) \(126\)
Step-by-step Solution
Detailed explanation
Let equation in new position is \((4 x-y+z-10)+\lambda(x+y-z-4)=0\) \(4(4+\lambda)-1 .(-1+\lambda)+1 .(1-\lambda)=0\) \(\Rightarrow \lambda=-9\) So equation in new position is \(-5 x-10 y+10 z+26=0\) \(\Rightarrow \alpha=\frac{54}{15}\)
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