JEE Mains · Maths · STD 12 - 11. three dimension geometry
The equation of the plane passing through the point \((1,2,-3)\) and perpendicular to the planes \(3 x+y-2 z=5\) and \(2 x-5 y-z=7,\) is
- A \(3 x-10 y-2 z+11=0\)
- B \(6 x-5 y-2 z-2=0\)
- C \(11 x+y+17 z+38=0\)
- D \(6 x-5 y+2 z+10=0\)
Answer & Solution
Correct Answer
(C) \(11 x+y+17 z+38=0\)
Step-by-step Solution
Detailed explanation
Normal vector : \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1\end{array}\right|=-11 \hat{i}-\hat{j}+17 \hat{k}\) So drs of normal to the required plane is \(<11,1,17>\) plane passes through \((1,2,-3)\) So eq \(^{n}\) of plane :…
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