JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the plane \(P : 8 x +\alpha_1 y +\alpha_2 z +12=0\) be parallel to the line \(L : \frac{ x +2}{2}=\frac{ y -3}{3}=\frac{ z +4}{5}\). If the intercept of \(P\) on the \(y\)-axis is 1 , then the distance between \(P\) and \(L\) is :
- A \(\sqrt{14}\)
- B \(\frac{6}{\sqrt{14}}\)
- C \(\sqrt{\frac{2}{7}}\)
- D \(\sqrt{\frac{7}{2}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{14}\)
Step-by-step Solution
Detailed explanation
P: \(8 x+\alpha_1 y+\alpha_2 z+12=0\) \(L : \frac{ x +2}{2}=\frac{ y -3}{3}=\frac{ z +4}{5}\) \(\because\) \(P\) is parallel to \(L\) \(\Rightarrow 8(2)+\alpha_1(3)+5\left(\alpha_2\right)=0\) \(\Rightarrow 3 \alpha_1+5\left(\alpha_2\right)=-16\) Also \(y\)-intercept of plane…
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