MHT CET · Maths · Three Dimensional Geometry
Equation of the plane passing through \((1,-1,2)\) and perpendicular to the planes \(x+2 y-2 z=4\) and \(3 x+2 y+z=6\) is
- A \(6 x-7 y-4 z-5=0\)
- B \(6 x+7 y-4 z+5=0\)
- C \(6 x-7 y+4 z+5=0\)
- D \(6 x+7 y+4 z-5=0\)
Answer & Solution
Correct Answer
(A) \(6 x-7 y-4 z-5=0\)
Step-by-step Solution
Detailed explanation
The equation of plane passing through \((1,-1,2)\) is \(\mathrm{a}(x-1)+\mathrm{b}(y+1)+\mathrm{c}(\mathrm{z}-2)=0\)
Since plane (i) is perpendicular to the planes \(x+2 y-2 z=4\) and \(3 x+2 y+z=6\)
\(\therefore a+2 b-2 c=0 \)
\( \text { and } 3 a+2 b+c=0 \)
\( \Rightarrow \frac{a}{6}=\frac{b}{-7}=\frac{c}{-4}\)
\(\therefore\) The equation of the required plane is
\(
\begin{aligned}
& 6(x-1)-7(y+1)-4(z-2)=0 \\
& \Rightarrow 6 x-7 y-4 z-5=0
\end{aligned}
\)
Since plane (i) is perpendicular to the planes \(x+2 y-2 z=4\) and \(3 x+2 y+z=6\)
\(\therefore a+2 b-2 c=0 \)
\( \text { and } 3 a+2 b+c=0 \)
\( \Rightarrow \frac{a}{6}=\frac{b}{-7}=\frac{c}{-4}\)
\(\therefore\) The equation of the required plane is
\(
\begin{aligned}
& 6(x-1)-7(y+1)-4(z-2)=0 \\
& \Rightarrow 6 x-7 y-4 z-5=0
\end{aligned}
\)
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