Q.47. If the line \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, then p =
A. \frac{70}{11}
B. \frac{11}{70}
C. \frac{-70}{11}
D. \frac{-11}{70}
Answer :- A. \frac{70}{11}
Explanation :-
Given lines can be written as
\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}
As these lines are at right angles,we get
(-3)\left ( -\frac{3p}{7} \right )+\left ( \frac{2p}{7} \right )(1)+(2)(-5)=0 \left ( \frac{9p}{7} \right )+\left ( \frac{2p}{7} \right )-10 = 0 \therefore \frac{11p}{7}=10 p=\frac{70}{11}