MHT CET · Maths · Three Dimensional Geometry
Equation of the plane, through the points \((-1,2,-2)\) and \((-1,3,2)\) and perpendicular to \(y z\) - plane, is
- A \(4 y+z=10\)
- B \(4 y-z+10=0\)
- C \(4 y-z=10\)
- D \(4 y+z+10=0\)
Answer & Solution
Correct Answer
(C) \(4 y-z=10\)
Step-by-step Solution
Detailed explanation
Equation of plane passing though \((-1,2,-2)\) and \((-1,3,2)\) is
\(\frac{x+1}{(-1+1)}=\frac{y-3}{2-3}=\frac{z-2}{-2-2}\)
Above plane is perpendicular to \(y \mathrm{z}\) - plane
\(\begin{aligned}
\therefore \quad & \frac{y-3}{-1}=\frac{z-2}{-4} \\
& \Rightarrow 4(y-3)=z-2 \\
& \Rightarrow 4 y-12-z+2=0 \\
& \Rightarrow 4 y-z=10
\end{aligned}\)
\(\frac{x+1}{(-1+1)}=\frac{y-3}{2-3}=\frac{z-2}{-2-2}\)
Above plane is perpendicular to \(y \mathrm{z}\) - plane
\(\begin{aligned}
\therefore \quad & \frac{y-3}{-1}=\frac{z-2}{-4} \\
& \Rightarrow 4(y-3)=z-2 \\
& \Rightarrow 4 y-12-z+2=0 \\
& \Rightarrow 4 y-z=10
\end{aligned}\)
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