JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the foot of perpendicular of the point \(P (3,-2,-9)\) on the plane passing through the points \((-1,-2,-3),(9,3,4),(9,-2,1)\) be \(Q(\alpha, \beta, \gamma)\). Then the distance of \(Q\) from the origin is:
- A \(\sqrt{29}\)
- B \(\sqrt{35}\)
- C \(\sqrt{42}\)
- D \(\sqrt{38}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{42}\)
Step-by-step Solution
Detailed explanation
\(P (3,-2,-9)\) Equation of plane through \(A,B,C.\) \(\left|\begin{array}{ccc}x+1 & y+2 & z+3 \\10 & 5 & 7 \\10 & 0 & 4\end{array}\right|=0\) \(2 x+3 y-5 z-7=0\) Foot of \(I ^{ r }\) of \(P (3,-2,-9)\) is \(\frac{x-3}{2}=\frac{y+2}{3}=\frac{z+9}{-5}=-\frac{(6-6+45-7)}{4+9+25}\)…
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