If 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, find the value of k.
If 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, find the value of k.
Solution and Explanation
The direction of ratios of 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 -3,2k,2 and 3k,1,-5 respectively.
It is known that two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular, if a1a2+b1b2+c1c2=0
∴-3(3k)+2k×1+2(-5)=0
⇒-9k+2k-10=0
⇒7k=-10
⇒k=-10/7
Therefore, for k=-10/7, the given lines are perpendicular to each other.
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