JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the foot of perpendicular from the point \(A (4,3\), 1) on the plane \(P: x-y+2 z+3=0\) be N. If \(B(5\), \(\alpha, \beta), \alpha, \beta \in Z\) is a point on plane \(P\) such that the area of the triangle \(A B N\) in \(3 \sqrt{2}\), then \(\alpha^2+\beta^2+\alpha \beta\) is equal to \(...........\).
- A \(6\)
- B \(5\)
- C \(7\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(7\)
Step-by-step Solution
Detailed explanation
\(AN =\sqrt{6}\) \(5-\alpha+2 \beta+3=0\) \(\alpha=8+2 \beta\) \(N\) is given by \(\frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}=\frac{-(4-3+2+3)}{1+1+4}\) \(x =3, y =4, z =-1\) \(N \text { is }(3,4,-1)\) \(BN =\sqrt{4+(\alpha-4)^2+(\beta+1)^2}\)…
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