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(D) Let the magnitudes of the vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = x$ and $|\overrightarrow{C}| = x\sqrt{2}$.Given $\overrightarr

Vedclass · Accepted Answer

$90^\circ, 135^\circ, 135^\circ$

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(D) Let the magnitudes of the vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = x$ and $|\overrightarrow{C}| = x\sqrt{2}$.Given $\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0$,which implies $\overrightarrow{A} + \overrightarrow{B} = -\overrightarrow{C}$.Squaring both sides: $|\overrightarrow{A}|^2 + |\overrightarrow{B}|^2 + 2|\overrightarrow{A}||\overrightarrow{B}|\cos(	heta_{AB}) = |\overrightarrow{C}|^2$.Substituting the values: $x^2 + x^2 + 2x^2\cos(	heta_{AB}) = (x\sqrt{2})^2 = 2x^2$.$2x^2 + 2x^2\cos(	heta_{AB}) = 2x^2$,which gives $\cos(	heta_{AB}) = 0$,so $	heta_{AB} = 90^\circ$.Since the vectors form a closed triangle,the exterior angles between the vectors are determined by the geometry of the triangle. The angles between the vectors are $90^\circ, 135^\circ, 135^\circ$.

Column $-I$	Column $-II$
$(a) \vec a + \vec b = \vec c$	$(i)$ Vector $\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ connects the origin to the tip of $\vec c$
$(b) \vec a - \vec c = \vec b$	$(ii)$ Vector $\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ connects the origin to the tip of $\vec b$
$(c) \vec b - \vec a = \vec c$	$(iii)$ Vector $\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$
$(d) \vec a + \vec b + \vec c = 0$	$(iv)$ Vector $\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ connects the origin to the tip of $\vec b$

Among three vectors,two vectors are equal in magnitude,and the magnitude of the third vector is $\sqrt{2}$ times the magnitude of the other two vectors. If $\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0$,find the angles between the vectors.

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