Question

If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at an angle $\theta$, then

• A. $|\vec{A} - \vec{B}| = \sqrt{2} R \sin \left(\frac{\theta}{2}\right)$
• B. $|\vec{A} + \vec{B}| = 2 R \sin \left(\frac{\theta}{2}\right)$
• C. $|\vec{A} + \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)$
• D. $|\vec{A} - \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)$

Answer 1

Answer: (C)

$|\vec{A} + \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)$

Answer 2

The magnitude of the resultant vector $R'$ of two vectors $A$ and $B$ inclined at an angle $\theta$ is given by:

$R' = \sqrt{a^2 + b^2 + 2ab \cos \theta}.$

Here $a = b = R$, so:

$R' = \sqrt{R^2 + R^2 + 2R \cdot R \cos \theta} = \sqrt{2R^2 (1 + \cos \theta)}.$

Using the identity $1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right)$, we get:

$R' = \sqrt{2R^2 \cdot 2 \cos^2 \left(\frac{\theta}{2}\right)} = 2R \cos \left(\frac{\theta}{2}\right).$

Thus, the answer is:

$|A + B| = 2R \cos \left(\frac{\theta}{2}\right).$