JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the lines \(\frac{x-k}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+3}{1}\) are co-planar, then the value of \(k\) is \(.....\)
- A \(5\)
- B \(4\)
- C \(1\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\(\left|\begin{array}{ccc}k+1 & 4 & 6 \\ 1 & 2 & 3 \\ 3 & 2 & 1\end{array}\right|=0\) \(\{\therefore\) Shortest distance between then is zero \(\}\) \((k+1)[2-6]-4[1-9]+6[2-6]=0\) \(\mathrm{~K}=1\)
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