# Let the angle between two non zero vectors A(vector) and B(vector ) be 120(degrees) and its resultant be C(vector). Then which of the following is(are) correct?

## (a)C must be equal to |A-B| (b)C must be less than |A-B| (c) C must be greater than |A-B| (d) C may be equal to |A-B|

Jun 12, 2018

Option (b)

#### Explanation:

$\boldsymbol{A} \cdot \boldsymbol{B} = \left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid \cos \left({120}^{o}\right) = - \frac{1}{2} \left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid$

$\boldsymbol{C} = \boldsymbol{A} + \boldsymbol{B}$

• ${C}^{2} = \left(\boldsymbol{A} + \boldsymbol{B}\right) \cdot \left(\boldsymbol{A} + \boldsymbol{B}\right)$

$= {A}^{2} + {B}^{2} + 2 \boldsymbol{A} \cdot \boldsymbol{B}$

$= {A}^{2} + {B}^{2} - \left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid q \quad \square$

• ${\left\mid \boldsymbol{A} - \boldsymbol{B} \right\mid}^{2}$

$= \left(\boldsymbol{A} - \boldsymbol{B}\right) \cdot \left(\boldsymbol{A} - \boldsymbol{B}\right)$

$= {A}^{2} + {B}^{2} - 2 \boldsymbol{A} \cdot \boldsymbol{B}$

$= {A}^{2} + {B}^{2} + \left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid q \quad \triangle$

${\left\mid \boldsymbol{A} - \boldsymbol{B} \right\mid}^{2} - {C}^{2} = \triangle - \square = 2 \left\mid \boldsymbol{A} \right\mid \left\mid \boldsymbol{B} \right\mid$

$\therefore {C}^{2} < {\left\mid \boldsymbol{A} - \boldsymbol{B} \right\mid}^{2} , q \quad \boldsymbol{A} , \boldsymbol{B} \ne \boldsymbol{0}$

$\therefore \left\mid \boldsymbol{C} \right\mid < \left\mid \boldsymbol{A} - \boldsymbol{B} \right\mid$