JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(d\) be the distance between the foot of perpendiculars of the points \(P (1,2-1)\) and \(Q (2,-\) 1,3 ) on the plane \(- x + y + z =1\). Then \(d ^{2}\) is equal to
- A \(16\)
- B \(36\)
- C \(26\)
- D \(46\)
Answer & Solution
Correct Answer
(C) \(26\)
Step-by-step Solution
Detailed explanation
Points \(P (1,2,-1)\) and \(Q (2,-1,3)\) lie on same side of the plane. Perpendicular distance of point \(P\) from plane is \(\left|\frac{-1+2-1-1}{\sqrt{1^{2}+1^{2}+1^{2}}}\right|=\frac{1}{\sqrt{3}}\) Perpendicular distance of point \(Q\) from plane is…
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