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

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 mrngBadiuddin broYour solution is wrongChk this one-Vector A=60i+3jVector B=40i-8jVector C=ai-52j   Find Vector AB ,BCVector AB=vector B-vector AVctr AB=-20i-11jVctr BC=(a-40)i-44j For collinear angle btwn vctr AB and vctr BC should be either 0 or 180 degreeThat means cross product of vctr AB and BC is zeroAB*BC(*for cross product)=(-20i-11j)*[(a-40)i-44j]AB*BC=[880+11(a-40)]k=0880+11(a-40)=080+a-40=0a=-40 answer
```
one year ago
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