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The value of \(x\) for which the lines \(\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\) and \(\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular is |
\(\frac{7}{10}\) \(\frac{10}{7}\) \(-\frac{10}{7}\) \(0\) |
\(-\frac{10}{7}\) |
Lines are perpendicular there for \(a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0\) ⇒ −3(3k)+2k×1+2(−5)=0 $ \Rightarrow 7k = -10$ $\Rightarrow k = \frac{-10}{7}$ |
