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  • If the lines x-1/-3 = y-2/k = z-3/2 and x-1/3k = y-1/1 = z-6/-5 are perpendicular, find the value of k.

6.  If the lines \frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2} and \frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}are perpendicular, find the value of k.

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Given both lines are perpendicular so we have the relation; a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

For the two lines whose direction ratios are known,

a_{1},b_{1},c_{1}\ and\ a_{2},b_{2},c_{2} 

We have the direction ratios of the lines, \frac{x-1}{-3}=\frac{y-2}{k}=\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.

Therefore applying the formula,

-3(3k)+2k(1)+2(-5) = 0

\Rightarrow -9k +2k -10 = 0

\Rightarrow7k=-10  or  k= \frac{-10}{7}

\therefore For, k= \frac{-10}{7} the lines are perpendicular.

 

Posted by

Divya Prakash Singh

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