JEE Mains · Maths · STD 12 - 11. three dimension geometry
Suppose the line \(\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}\) lies on the plane \(x+3 y-2 z+\beta=0 .\) Then \((\alpha+\beta)\) is equal to ... .
- A \(5\)
- B \(7\)
- C \(6\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
Point \((2,2,-2)\) also lies on given plane So \(2+3 \times 2-2(-2)+\beta=0\) \(\Rightarrow 2+6+4+\beta=0 \Rightarrow \beta=-12\) Also \(\alpha \times 1-5 \times 3+2 \times-2=0\) \(\Rightarrow \alpha-15-4=0 \Rightarrow \alpha=19\) \(\therefore \alpha+\beta=19-12=7\)
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