Question 40: Motion in 2D & Vectors

Question

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

Accepted Answer

The magnitude of resultant vector
$R^{\prime}=\sqrt{a^2+b^2+2 a b \cos 	heta}$
Here $a=b=R$
Then $R^{\prime}=\sqrt{R^2+R^2+2 R^2 \cos 	heta}$
$\begin{aligned}
& =R \sqrt{2} \sqrt{1+\cos 	heta} \
& =\sqrt{2} R \sqrt{2 \cos ^2 \frac{	heta}{2}} \
& =2 R \cos \frac{	heta}{2}
\end{aligned}$

Answer

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

Answer

$|\vec{A}-\vec{B}|=2 R \cos \left(\frac{	heta}{2}ight)$

Answer

$|\vec{A}-\vec{B}|=\sqrt{2} R \sin \left(\frac{	heta}{2}ight)$

Answer

$|\vec{A}+\vec{B}|=2 R \sin \left(\frac{	heta}{2}ight)$

Motion in 2D & Vectors

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

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