JEE Mains · Maths · STD 12 - 11. three dimension geometry
If two distinct point \(Q , R\) lie on the line of intersection of the planes \(- x +2 y - z =0\) and \(3 x-5 y+2 z=0\) and \(P Q=P R=\sqrt{18}\) where the point \(P\) is \((1,-2,3)\), then the area of the triangle \(PQR\) is equal to
- A \(\frac{2}{3} \sqrt{38}\)
- B \(\frac{4}{3} \sqrt{38}\)
- C \(\frac{8}{3} \sqrt{38}\)
- D \(\sqrt{\frac{152}{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{3} \sqrt{38}\)
Step-by-step Solution
Detailed explanation
\(-x+2 y-z=0\) \(3 x-5 y+2 z=0\) \(\overrightarrow{ n }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ -1 & 2 & -1 \\ 3 & -5 & 2\end{array}\right|\) \(=\hat{ i }(-1)-\hat{ j }(1)+\hat{ k }(-1)\) \(\overrightarrow{ n }=-\hat{ i }-\hat{ j }-\hat{ k }\) Equation of…
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