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  • Find the vector and the cartesian equations of the through the point ( 5 , 2, -4) and which is parallel to the vector 3i+2j-8k .

Find the vector and the cartesian equations of the through the point ( 5 , 2, -4) and which is parallel to the vector 3i+2j-8k .

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 Solution:  We have , 

                                        veca= 5hati+2hatj-4hatk    and   vecb= 3hati+2hatj-8hatk 

               Therefore , the vector equation of the line is 

                                        vecr=5hati+2hatj-4hatk+lambda(3hati+2hatj-8hatk)

              Now , vecr is the position vector of any point P(x,y,z) on the line .

             Therefore,           xhati+yhatj+zhatk=5hati+2hatj-4hatk+lambda(3hati+2hatj-8hatk)

                                                                       =(5+3lambda)hati+(2+2lambda)hatj+(-4-8lambda)hatk

Eliminating lambda , we get 

                                   fracx-53=fracy-22=fracz+4-8

which is the equation of the line in cartesian form .   

Posted by

Deependra Verma

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