JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the point \((7,10,11)\) from the line \(\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}\) along the line \(\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}\) is
- A \(18\)
- B \(14\)
- C \(12\)
- D \(16\)
Answer & Solution
Correct Answer
(B) \(14\)
Step-by-step Solution
Detailed explanation
\(\because\) line \(P Q\) is parallel to line \(\frac{x-9}{2}=\frac{y-3}{3}=\frac{z-17}{6}\) \(\begin{aligned} & \therefore \frac{\lambda-3}{2}=\frac{-6}{3}=\frac{3 \lambda-9}{6} \Rightarrow \lambda=-1 \\ & Q=(3,4,-1) \\ & \therefore P Q=\sqrt{16+36+144}=14 \end{aligned}\)
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