JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of line \(3 y-2 z-1=0=3 x-z+4\) from the point \((2,-1,6)\) is :
- A \(\sqrt{26}\)
- B \(2 \sqrt{5}\)
- C \(2 \sqrt{6}\)
- D \(4 \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
\(3 y-2 z-1=0=3 x-z+4\) \(3 y-2 z-1=0\) D.R's \(\Rightarrow(0,3,-2)\) \(3 x-z+4=0\) D. R's \(\Rightarrow(3,-1,0)\) Let DR's of given line are \(a, b, c\) Now \(3 \mathrm{~b}-2 \mathrm{c}=0 \,\& 3 \mathrm{a}-\mathrm{c}=0\) \(\therefore 6 \mathrm{a}=3 \mathrm{~b}=2 \mathrm{c}\)…
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