Question

Two vectors A and B have magnitudes 6 units and 8 units respectively. Find |A-B|, if the angle between two vectors is .(a) 0^(@) (b) 180^(@) (c) 180^(@) (d) 120^(@) .

Answer 1

(a):2

(B):14

(C)14

(D)12.16

Answer 2

Magnitude of vector A = 6 units.

Magnitude of vector B = 8 units.

Now |A-B| will be equal to $\sqrt{A^{2}+B^{2}-2*A*BCos(angle) }$.

1. Case 1 : when angle between them is 0 degree:

 then |A-B|=$\sqrt{6^{2}+8^{2}-2*6*8*Cos(0) }$

                 =$\sqrt{100-96}$  ( Since Cos(0)=1)

                =2 Units.

2. Case 2: When angle between them is  180 degree:

then |A-B|  = $\sqrt{6^{2}+8^{2}-2*6*8*Cos(180) }$

                  =$\sqrt{100+96}$                (Since Cos(180) is -1)

                  = 14 Units.

3. Case 3 : Case 3 is same as case 2. So the answer is same.

4. Case 4: When the angle between them is 120 degree.

         then |A-B|  =$\sqrt{6^{2}+8^{2}-2*6*8*Cos(120) }$

                           =$\sqrt{100+48}$         (Since Cos(120)= -1/2)

                           =12.16 Units.