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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 2
i wanted the solution also
8 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
```
8 years ago
```							thanks bro
```
8 years ago
```							sq root of 3
```
7 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
```							Option a is the correct as by solving resultant we get cos theta as -1/2 thus substitute value in a-b resultant
```
one year 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=D²So the magnitude of their difference will be  ROOT 3
```
one year ago
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