 # The points with position vectors 60i+30j,40i-8j,ai-52j are collinear if a=  Badiuddin askIITians.ismu Expert
148 Points
12 years ago

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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5 years ago
Hello bro gud mrng
Chk this one-
Vector A=60i+3j
Vector B=40i-8j
Vector C=ai-52j
Find Vector AB ,BC
Vector AB=vector B-vector A
Vctr AB=-20i-11j
Vctr BC=(a-40)i-44j
For collinear angle btwn vctr AB and vctr BC should be either 0 or 180 degree
That means cross product of vctr AB and BC is zero
AB*BC(*for cross product)
=(-20i-11j)*[(a-40)i-44j]
AB*BC=[880+11(a-40)]k=0
880+11(a-40)=0
80+a-40=0