Find the value of p for which the following lines are perpendicular:

\frac{1-x}{3}=\frac{2y-14}{2p}=\frac{z-3}{2};\: \frac{1-x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}

 

 

 

 

 
 
 
 
 

Answers (1)

\frac{1-x}{3}=\frac{2y-14}{2p}=\frac{z-3}{2}

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

\frac{x-1}{-3}=\frac{y-7}{p}=\frac{z-3}{2}........(1)

\frac{x-1}{-3p}=\frac{y-5}{1}=\frac{z-6}{-5}........(2)

D.R of (1) (-3,p,2)   D.R of (2) (-3p,1,-5)

DR_1.DR_2\Rightarrow (-3)(-3p)+(p)(1)+2(-5)=0

\Rightarrow 9p+p-10=0

\Rightarrow 10p=10

\Rightarrow p=1

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