MHT CET · Maths · Three Dimensional Geometry
The line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane \(x+3 y-\alpha z+\beta=0\), then value of \(\alpha \beta\) is
- A 42
- B 1
- C -42
- D -2
Answer & Solution
Correct Answer
(C) -42
Step-by-step Solution
Detailed explanation
line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane \(x+3 y-\alpha z+\beta=0\)
\(\therefore(3)(1)+(-5)(3)+2(-\alpha)=0 \Rightarrow \alpha=-6\)
Thus equation of plane is \(x+3 y+6 z+\beta=0\) and point \((2,1,-2)\) lies in it.
\(
\begin{aligned}
& \therefore 2+3(1)+6(-2)+\beta=0 \Rightarrow \beta=7 \\
& \therefore \alpha \beta=(-6)(7)=-42
\end{aligned}
\)
\(\therefore(3)(1)+(-5)(3)+2(-\alpha)=0 \Rightarrow \alpha=-6\)
Thus equation of plane is \(x+3 y+6 z+\beta=0\) and point \((2,1,-2)\) lies in it.
\(
\begin{aligned}
& \therefore 2+3(1)+6(-2)+\beta=0 \Rightarrow \beta=7 \\
& \therefore \alpha \beta=(-6)(7)=-42
\end{aligned}
\)
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