Question

Let the angle between two nonzero vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ be 120° and its resultant be $\overrightarrow{C}$.
(a) C must be equal to $\left|A-B\right|$
(b) C must be less than $\left|A-B\right|$
(c) C must be greater than $\left|A-B\right|$
(d) C may be equal to $\left|A-B\right|$

Answer 1

(b) C must be less than $\left|A-B\right|$

Here, we have three vector A, B and C.
$$\begin{gathered} {\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2={\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2+2\overrightarrow{A}.\overrightarrow{B}...\left(i\right) \\ {\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2={\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2-2\overrightarrow{A}.\overrightarrow{B}...\left(ii\right) \end{gathered}$$

Subtracting (i) from (ii), we get:
${\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2-{\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2=4\overrightarrow{A}.\overrightarrow{B}$

Using the resultant property $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$, we get:
$$\begin{gathered} {\left|\overrightarrow{C}\right|}^2-{\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2=4\overrightarrow{A}.\overrightarrow{B} \\ \Rightarrow {\left|\overrightarrow{C}\right|}^2={\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2+4\overrightarrow{A}.\overrightarrow{B} \\ \Rightarrow {\left|\overrightarrow{C}\right|}^2={\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2+4\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos 120° \end{gathered}$$

Since cosine is negative in the second quadrant, $C$must be less than $\left|A-B\right|$.