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(B) Let the vectors be $\vec{OA}, \vec{OB}, \vec{OC}$. We need to find the magnitude of $\vec{R} = \vec{OA} + \vec{OB} + \vec{OC}$.Let $\vec{OB}$ be a

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$(\sqrt{2}+1) R$

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(B) Let the vectors be $\vec{OA}, \vec{OB}, \vec{OC}$. We need to find the magnitude of $\vec{R} = \vec{OA} + \vec{OB} + \vec{OC}$.Let $\vec{OB}$ be along the $y$-axis,so $\vec{OB} = R\hat{j}$.Then $\vec{OA} = R(\cos 45^{\circ} \hat{i} + \sin 45^{\circ} \hat{j}) = R(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j})$.And $\vec{OC} = R(\cos 45^{\circ} \hat{i} - \sin 45^{\circ} \hat{j}) = R(\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j})$.Summing these vectors:$\vec{R} = \vec{OA} + \vec{OB} + \vec{OC} = R(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}) + R\hat{j} + R(\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j})$$\vec{R} = R(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})\hat{i} + R(\frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}})\hat{j}$$\vec{R} = R(\frac{2}{\sqrt{2}})\hat{i} + R\hat{j} = \sqrt{2}R\hat{i} + R\hat{j}$.The magnitude is $|\vec{R}| = \sqrt{(\sqrt{2}R)^2 + R^2} = \sqrt{2R^2 + R^2} = \sqrt{3R^2} = \sqrt{3}R$.Wait,re-evaluating the geometry: The resultant of $\vec{OA}$ and $\vec{OC}$ is $2R \cos(45^{\circ}) \hat{j} = \sqrt{2}R \hat{j}$.Adding $\vec{OB} = R \hat{j}$,the total resultant is $(\sqrt{2} + 1)R \hat{j}$.Thus,the magnitude is $(\sqrt{2} + 1)R$.

Let $A, B$ and $C$ be three points on a circle of radius $R$. If $O$ is the centre of the circle and $\angle AOB = 45^{\circ}, \angle BOC = 45^{\circ}$,then the resultant of $\vec{OA}, \vec{OB}$ and $\vec{OC}$ has magnitude

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