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Let P(3,2,6) be a point in space and Q be a point on line Vector r =(i-j+2k)+mu (-3i+j+5k) .then the value of mu which the vector is PQ parallel to the plane x-4y+3z=1 is

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Solution:    Any point  on the line can be taken as 

                    \ \ Q=(1-3mu , mu-1,5mu+2)\ \ vecPQ=(-3mu-2,mu-3,5mu-4)

Now             1(-3mu-2)-4(mu-3)+3(5mu-4)=0

          \ \ Rightarrow hspace1cm8mu=2 Rightarrow mu=frac14

Posted by

Deependra Verma

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