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pallavi pradeep bhardwaj Grade: 12
        

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

7 years ago

Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points
										

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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7 years ago
Jitender singh
11 Points
										
Hello bro gud mrng
Badiuddin bro
Your solution is wrong
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
a=-40 answer
3 months ago
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