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Physics and Mathematics — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsPhysics and Mathematics1 Mark
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 $|\text{A}-\text{B}|$
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 $|\text{A}-\text{B}|.$

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