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Find the values of p so that the lines 1-x/ 3 =7 y- 14/ 2 p = z- 3/2 and 7 - 7 x /3 p = y-5/ 1 = 6-z /5 are at right angles.

12.   Find the values of p so that the lines \frac{1-x}{3}=\frac{7y-14}{2p}= \frac{z-3}{2} and \frac{7-7x}{3p}=\frac{y-5}{1}= \frac{6-z}{5} are at right angles.

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First we have to write the given equation of lines in the standard form;

\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}= \frac{z-3}{2}   and  \frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}= \frac{z-6}{-5}

Then we have the direction ratios of the above lines as;

-3,\ \frac{2p}{7},\ 2   and   \frac{-3p}{7},\ 1,\ -5   respectively..

Two lines with direction ratios a_{1},b_{1},c_{1} and a_{2},b_{2},c_{2}  are perpendicular to each other if,  a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0

\therefore (-3).\left ( \frac{-3p}{7} \right )+(\frac{2p}{7}).(1) + 2.(-5) = 0

\Rightarrow \frac{9p}{7}+ \frac{2p}{7} =10

\Rightarrow 11p =70

\Rightarrow p =\frac{70}{11}

Thus, the value of p is \frac{70}{11}.

 

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