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Let $A, B$ and $C$ is three points on a circle of radius $R$, if $O$ is the centre of the circle and $\angle A O B=45^{\circ}, \angle B O C=45^{\circ}$, then the resultant of $\mathbf{O A}, \mathbf{O B}$ and $\mathbf{O C}$ has magnitude
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Verified Answer
The correct answer is:
$(\sqrt{2}+1) R$
Given, diagram can be projected on $x y-$ plane as shown here.
$|\mathbf{O A}|=R=|\mathbf{O B}|=|\mathbf{O C}|$
$\mathbf{O B}$ has projected as $\mathbf{A} \mathbf{B}^{\prime}$ and $\mathbf{O C}$ is projected as $\mathbf{B}^{\prime} \mathbf{C}^{\prime}$.
Hence, $\left|\mathbf{A B}^{\prime}\right|=\left|\mathbf{B}^{\prime} \mathbf{C}^{\prime}\right|=R$ and $\angle \mathbf{B B}^{\prime} \mathbf{C}^{\prime}=90^{\circ}$.
Hence, the resultant of $\mathbf{O A}, \mathbf{O B}$ and $\mathbf{O C}$ is $\mathbf{O C}^{\prime}$
$\mathbf{O C}^{\prime}=\mathbf{O B}+\mathbf{B C}^{\prime}$
$\Rightarrow \quad\left|\mathbf{O C}^{\prime}\right|=|\mathbf{O B}|+\left|\mathbf{B C}^{\prime}\right|$ ...(i)
In $\triangle B B^{\prime} C^{\prime}$, using Pythagoras
$\left|\mathbf{B C}^{\prime}\right|=\sqrt{R^2+R^2}=\sqrt{2} R$
From Eq. (i), we get
$\left|\mathbf{O C}^{\prime}\right|=R+\sqrt{2} R$
$=(\sqrt{2}+1) R$
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