MHT CET · Maths · Three Dimensional Geometry
If 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 \((\alpha, \beta)=\)
- A \((6,-7)\)
- B \((-6,7)\)
- C \((5,-15)\)
- D \((-5,15)\)
Answer & Solution
Correct Answer
(B) \((-6,7)\)
Step-by-step Solution
Detailed explanation
Given equation of line
\(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\)
\(\therefore \quad\) The line passes through \((2,1,-2)\)
The above point lies on the plane
\(\begin{aligned}
& x+3 y-\alpha z+\beta=0 \\
& \Rightarrow 2+3+2 \alpha+\beta=0 \\
& \Rightarrow 2 \alpha+\beta=-5
\end{aligned}\)
Also the given line is perpendicular to the normal to the plane
\(\begin{aligned}
& a_1 a_2+b_1 b_2+c_1 c_2=0 \\
& \Rightarrow 3(1)+(-5)(3)-2(\alpha)=0 \\
& \Rightarrow \alpha=-6
\end{aligned}\)
\(\begin{aligned} & \text { From (i) } \\ & \beta=7 \\ \therefore \quad & \alpha \beta=-42 \\ & (\alpha, \beta)=(-6,7)\end{aligned}\)
\(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\)
\(\therefore \quad\) The line passes through \((2,1,-2)\)
The above point lies on the plane
\(\begin{aligned}
& x+3 y-\alpha z+\beta=0 \\
& \Rightarrow 2+3+2 \alpha+\beta=0 \\
& \Rightarrow 2 \alpha+\beta=-5
\end{aligned}\)
Also the given line is perpendicular to the normal to the plane
\(\begin{aligned}
& a_1 a_2+b_1 b_2+c_1 c_2=0 \\
& \Rightarrow 3(1)+(-5)(3)-2(\alpha)=0 \\
& \Rightarrow \alpha=-6
\end{aligned}\)
\(\begin{aligned} & \text { From (i) } \\ & \beta=7 \\ \therefore \quad & \alpha \beta=-42 \\ & (\alpha, \beta)=(-6,7)\end{aligned}\)
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