JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P\) be the plane, passing through the point \((1,-1,-5)\) and perpendicular to the line joining the points \((4,1,-3)\) and \((2,4,3)\). Then the distance of \(P\) from the point \((3,-2,2)\) is
- A \(6\)
- B \(4\)
- C \(5\)
- D \(7\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
Equation of Plane : \(2(x-1)-3(y+1)-6(z+5)=0\) Or \(2 x-3 y-6 z=35\) \(\Rightarrow\) Re quired distance \(=\) \(\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}\) \(=5\)
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