Two vectors overrightarrow{a} and overrightar | vedclass

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(B) Let $	heta = 60^{\circ}$ be the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$. Let $\alpha = 45^{\circ}$ be the angle between the r

Vedclass · Accepted Answer

$\sqrt{3} - 1$

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(B) Let $	heta = 60^{\circ}$ be the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$. Let $\alpha = 45^{\circ}$ be the angle between the resultant $\overrightarrow{R}$ and $\overrightarrow{a}$.Using the formula for the direction of the resultant vector: $	an \alpha = \frac{b \sin 	heta}{a + b \cos 	heta}$.Substituting the given values: $	an 45^{\circ} = \frac{2 \sin 60^{\circ}}{a + 2 \cos 60^{\circ}}$.Since $	an 45^{\circ} = 1$,$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,and $\cos 60^{\circ} = \frac{1}{2}$,we have:$1 = \frac{2(\frac{\sqrt{3}}{2})}{a + 2(\frac{1}{2})}$$1 = \frac{\sqrt{3}}{a + 1}$$a + 1 = \sqrt{3}$$a = \sqrt{3} - 1$.

Two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are at an angle of $60^{\circ}$ with each other. Their resultant makes an angle of $45^{\circ}$ with $\overrightarrow{a}$. If $|\overrightarrow{b}| = 2 \text{ units}$,then $|\overrightarrow{a}|$ is:

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