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JEE Mains · Maths · STD 12 - 11. three dimension geometry

બિંદુ \((7,10,11)\) નું રેખા \(\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}\) થી રેખા \(\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}\) પર અંતર ___ છે.

  1. A \(18\)
  2. B \(14\)
  3. C \(12\)
  4. D \(16\)
Verified Solution

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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