Let the angle between two nonzero vectors A and B be 120° and its resultant be c : 1. C must be equal to |A-B| 2. C must be less than |A-B| 3. C must be greater than |A-B| 4. C may be equal to |A-B|

2. C must be less than |A-B| Explanation: Here, we have three vector A, B and C. | A + B |^2= | A |^2+ | B |^2+2 A . B ... (i) | A - B |^2= | A |^2+ | B |^2-2 A . B ... (ii) Subtracting (i) from (ii), we get: | A + B |^2- | A - B |^2=4 A . B Using the resultant property C = A + B , we get: | C |^2- | A - B |^2=4 A . B | C |^2= | A - B |^2+4 A . B | C |^2= | A - B |^2+4 | A |. | B | 120^ Since cosine is negative in the second quadrant, C must be less than |A-B|.

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

Download App

Question

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

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

✓

Answer

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

Explanation:
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 $|\text{A}-\text{B}|.$