JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance, of the point \((7,-2,11)\) from the line \(\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}\) along the line \(\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}\), is :
- A \(12\)
- B \(14\)
- C \(18\)
- D \(21\)
Answer & Solution
Correct Answer
(B) \(14\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)\) Point \(B\) lies on \(\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}\) \(\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}\) \(-3 \lambda-6=0\) \(\lambda=-2\) \(\text { B } \Rightarrow(3,4,-1)\)…
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