Let the angle between two nonzero vectors A and B be 120° and its…

MCQ

Let the angle between two nonzero vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ be 120° and its resultant be $\vec{\text{c}}:$

A
C must be equal to $|\text{A}-\text{B}|$
✓
C must be less than $|\text{A}-\text{B}|$
C
C must be greater than $|\text{A}-\text{B}|$
D
C may be equal to $|\text{A}-\text{B}|$

✓

Answer

Correct option: B.

C must be less than $|\text{A}-\text{B}|$

Here, we have three vector A, B and C.
$\big|\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\big|^2=\big|\overrightarrow{\text{A}}\big|^2+\big|\overrightarrow{\text{B}}\big|^2+2\overrightarrow{\text{A}}.\overrightarrow{\text{B}} \ ...{\text{(i)}}$
$\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2=\big|\overrightarrow{\text{A}}\big|^2+\big|\overrightarrow{\text{B}}\big|^2-2\overrightarrow{\text{A}}.\overrightarrow{\text{B}} \ ...{\text{(ii)}}$
Subtracting (i) from (ii), we get:
$\big|\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\big|^2-\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2=4\overrightarrow{\text{A}}.\overrightarrow{\text{B}}$
Using the resultant property $\overrightarrow{\text{C}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}},$ we get:
$\big|\overrightarrow{\text{C}}\big|^2-\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2=4\overrightarrow{\text{A}}.\overrightarrow{\text{B}}$
$\Rightarrow\big|\overrightarrow{\text{C}}\big|^2=\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2+4\overrightarrow{\text{A}}.\overrightarrow{\text{B}}$
$\Rightarrow\big|\overrightarrow{\text{C}}\big|^2=\big|\overrightarrow{\text{A}}-\overrightarrow{\text{B}}\big|^2+4\big|\overrightarrow{\text{A}}\big|.\big|\overrightarrow{\text{B}}\big|\cos120^{\circ}$
Since cosine is negative in the second quadrant, C must be less than $|\text{A}-\text{B}|.$