The points with position vectors 60i+30j,40i-8j,ai-52j are collinear if a=

Dear pallavi

if vector are collinier 

from 3 point form two vector

angel between these two vector is zero ,so cross product will be zero

 let two vector  a= (40i-8j) - (60i+30j)  

                         = -20 i -38 j

 and  b= (ai-52j ) - ( 60i+30j)

            =(a-60) i -82j

now aXb=O

(-20 i -38 j) X ((a-60) i -82j) =O

 1640 k  +38(a-60)k =O

 1640 +38 a - 2280  =0

a =640/38

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