JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P _1\) be the plane \(3 x - y -7 z =11\) and \(P _2\) be the plane passing through the points \((2,-1,0)\), \((2,0,-1)\), and \((5,1,1)\). If the foot of the perpendicular drawn from the point \((7,4,-1)\) on the line of intersection of the planes \(P_1\) and \(P_2\) is \((\alpha, \beta\), \(\gamma\) ), then \(\alpha+\beta+\gamma\) is equal to \(............\).
- A \(11\)
- B \(10\)
- C \(12\)
- D \(14\)
Answer & Solution
Correct Answer
(A) \(11\)
Step-by-step Solution
Detailed explanation
\(P_2\) is given by \(\left|\begin{array}{ccc}x-5 & y-1 & z-1 \\ 3 & 2 & 1 \\ 3 & 1 & 2\end{array}\right|=0\) \(x-y-z=3\) DR of line intersection of \(P _1\) and \(P _2\) \(\left|\begin{array}{ccc}i & j & k \\ 1 & -1 & 1 \\ 3 & -1 & -7\end{array}\right|\)…
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