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# if the sum of two unit vectors is a unit vector, then the magnitude of their difference is(a) sq root of 3(b) sq root of 2(c) sq root of 5(d) 1/sq root of 2i wanted the solution alsowaiting for ur reply

## 6 Answers

10 years ago

let two vectors are a,b then

resultant of sum = (a2 + b2 + 2abcos@)1/2 = 1               (given)

a = b = 1          (unit vector)

1+1+2cos@ = 1

cos@ = -1/2           .................1

now resultant of difference = magnitude (a-b)

R = (a2 + b2 - 2abcos@)1/2

a = b = 1 , cos@ = -1/2 so

R = [ 1+1+1]1/2 = root3

option a) is correct

3 years ago
let two vectors are a,b then
resultant of sum = (a2 + b2 + 2abcos@)1/2 = 1               (given)

a = b = 1          (unit vector)

1+1+2cos@ = 1

cos@ = -1/2           .................1

now resultant of difference = magnitude (a-b)

R = (a2 + b2 - 2abcos@)1/2

a = b = 1 , cos@ = -1/2 so

R = [ 1+1+1]1/2 = root3

option a) is correct

2 years ago
If A and B are two unit vectors and their resultant is as well a unit vector so the magnitude of A= that of B= that of R .
For this condition to satisfy the angle between vectors A and B should be 120° .
So, (A²+B²+2AB cos120°)=R²=1
Cos120° =[cos(90×1+30)]= – sin30°.         {using Alloy angle tool}
So we get : 1²+1²+2×-½=1 this proved that 120° is angle between them .
So now difference can be found by:
(A²+B² – 2AB cos120°)= D².                                  {D= difference magnitude }
So 1+1 – 2×(- ½) = D²
So , 1+1+1=D²
So the magnitude of their difference will be  ROOT 3

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