JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the lines \(\frac{{x - 2}}{1} = \frac{{y - 3}}{1} = \frac{{z - 4}}{{ - k}}\) and \(\frac{{x - 1}}{k} = \frac{{y - 4}}{2} = \frac{{z - 5}}{1}\) are coplanar then \(k \) can have
- A any value
- B exactly one value
- C exactly two value
- D exactly three value
Answer & Solution
Correct Answer
(C) exactly two value
Step-by-step Solution
Detailed explanation
\([a-c, b, d]=0\) \(\left| {\begin{array}{*{20}{c}} {2 - 1}&{3 - 4}&{4 - 5}\\ 1&1&{ - k}\\ k&2&1 \end{array}} \right| = 0\) \(\left| {\begin{array}{*{20}{c}} 1&{ - 1}&{ - 1}\\ 1&1&{ - k}\\ k&2&1 \end{array}} \right| = 0\) \(\Rightarrow 1(1+2 k)+\left(1+k^{2}\right)-(2-k)=0\)…
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