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JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the lines \(\frac{{x + 1}}{2} = \frac{{y - 1}}{1} = \frac{{z + 1}}{3}\) and \(\frac{{x + 2}}{2} = \frac{{y - k}}{3} = \frac{z}{4}\) are coplanar, then the value of \(k\) is
- A \(\frac{{11}}{2}\)
- B \(-\frac{{11}}{2}\)
- C \(\frac{{9}}{2}\)
- D \(-\frac{{9}}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{{11}}{2}\)
Step-by-step Solution
Detailed explanation
Two given planes are coplanar, if \(\left| {\begin{array}{*{20}{c}} { - 2 - \left( { - 1} \right)}&{k - 1}&{0 - \left( { - 1} \right)}\\ 2&1&3\\ 2&3&4 \end{array}} \right| = 0\)…
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