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(B) The angle between the vectors is $	heta = 60^{\circ}$.The angle between the resultant vector and $\vec{a}$ is $\phi = 45^{\circ}$.According to th

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$\sqrt{3}-1$

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(B) The angle between the vectors is $	heta = 60^{\circ}$.The angle between the resultant vector and $\vec{a}$ is $\phi = 45^{\circ}$.According to the parallelogram law of vector addition,the angle $\phi$ that the resultant makes with vector $\vec{a}$ is given by:$	an(\phi) = \frac{|\vec{b}| \sin(	heta)}{|\vec{a}| + |\vec{b}| \cos(	heta)}$Substituting the given values $|\vec{b}| = 2$,$	heta = 60^{\circ}$,and $\phi = 45^{\circ}$:$	an(45^{\circ}) = \frac{2 \sin(60^{\circ})}{|\vec{a}| + 2 \cos(60^{\circ})}$Since $	an(45^{\circ}) = 1$,$\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$,and $\cos(60^{\circ}) = \frac{1}{2}$:$1 = \frac{2 	imes (\frac{\sqrt{3}}{2})}{|\vec{a}| + 2 	imes (\frac{1}{2})}$$1 = \frac{\sqrt{3}}{|\vec{a}| + 1}$$|\vec{a}| + 1 = \sqrt{3}$$|\vec{a}| = \sqrt{3} - 1$ units.

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 $|\vec{b}|=2$ units,then $|\vec{a}|$ is

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