JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the line \(\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}\) from the point \((1,4,0)\) along the line \(\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}\) is :
- A \(\sqrt{17}\)
- B \(\sqrt{15}\)
- C \(\sqrt{14}\)
- D \(\sqrt{13}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{14}\)
Step-by-step Solution
Detailed explanation
Line passing through \((1,4,0)\) and parallel to \(\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}\) is \(L: \frac{x-1}{1}=\frac{y-4}{2}=\frac{z}{3}\) Any point on \(L:(\lambda+1,2 \lambda+4,3 \lambda)\) Any point on \(\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}\) is…
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