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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry

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

1 Answer

DRS of both lines is equal to 0.

(3,2p,2)•(3p,1,5)=0

11p+10=0

p=-10/11

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answered Jan 23 by adarsh06032000
 

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