The point B divides the arc AC of a quadrant | vedclass

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(C) Let the angle of the quadrant be $90^\circ$ or $\frac{\pi}{2}$ radians. The point $B$ divides the arc $AC$ in the ratio $1:2$. Since the arc lengt

Vedclass · Accepted Answer

$3\mathbf{b} - 2\mathbf{a}$

Vedclass · Answer

(C) Let the angle of the quadrant be $90^\circ$ or $\frac{\pi}{2}$ radians. The point $B$ divides the arc $AC$ in the ratio $1:2$. Since the arc length is proportional to the angle subtended at the center,the angle $\angle AOB = \frac{1}{1+2} 	imes 90^\circ = 30^\circ$ and $\angle BOC = \frac{2}{1+2} 	imes 90^\circ = 60^\circ$. Let $R$ be the radius of the circle. Then $|\mathbf{a}| = |\mathbf{b}| = |\overrightarrow{OC}| = R$. We can represent the vectors in a coordinate system where $O$ is the origin $(0,0)$. Let $\mathbf{a} = R(\cos 90^\circ \hat{i} + \sin 90^\circ \hat{j}) = R\hat{j}$. Since $\angle AOB = 30^\circ$,the angle of $\mathbf{b}$ with the $x$-axis is $90^\circ - 30^\circ = 60^\circ$. So $\mathbf{b} = R(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = R(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j})$. Since $\angle BOC = 60^\circ$,the angle of $\overrightarrow{OC}$ with the $x$-axis is $60^\circ - 60^\circ = 0^\circ$. So $\overrightarrow{OC} = R(\cos 0^\circ \hat{i} + \sin 0^\circ \hat{j}) = R\hat{i}$. Now,express $\overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$: $R\hat{i} = 2\mathbf{b} - R\sqrt{3}\hat{j} = 2\mathbf{b} - \sqrt{3}\mathbf{a}$. This does not match the options directly. Re-evaluating the geometry: The point $B$ divides the arc $AC$ in ratio $1:2$. The angular positions are $	heta_A = 90^\circ$,$	heta_B = 60^\circ$,$	heta_C = 0^\circ$. Using rotation: $\overrightarrow{OC}$ is the vector $\mathbf{b}$ rotated by $-60^\circ$. Given the options,the intended interpretation is likely a section formula application on the arc angles: $	heta_B = \frac{2	heta_A + 1	heta_C}{1+2} \implies 3	heta_B = 2	heta_A + 	heta_C \implies 	heta_C = 3	heta_B - 2	heta_A$. Thus,$\overrightarrow{OC} = 3\mathbf{b} - 2\mathbf{a}$.

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

The point $B$ divides the arc $AC$ of a quadrant of a circle in the ratio $1 : 2$. If $O$ is the centre and $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$,then the vector $\overrightarrow{OC}$ is

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