Question

Given that $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=0$ , out of three vectors two are equal in magnitude and the magnitude of third vector is $\sqrt{2}$ times that of either of two having equal magnitude. Then angle between vectors are given by

• A. ${30}^{\circ },{60}^{\circ },{90}^{\circ }$
• B. ${45}^{\circ },{135}^{\circ },{150}^{\circ }$
• C. ${90}^{\circ },{135}^{\circ },{150}^{\circ }$
• D. ${90}^{\circ },{135}^{\circ },{135}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ },{135}^{\circ },{135}^{\circ }$

$\begin{array}{c}Appling\sin eruleinthetriangle,\\ \frac{A}{\sin \alpha }=\frac{B}{\sin \beta }=\frac{C}{\sin \gamma }\\ or,\frac{A}{\sin \alpha }=\frac{C}{\sin \gamma }\\ or,\frac{A}{\sin \alpha }=\frac{A\sqrt{2}}{\sin \left\{{180}^0-\left(\alpha +\alpha \right)\right\}}\\ or,\frac{1}{\sin \alpha }=\frac{\sqrt{2}}{\sin 2\alpha }\\ or,\frac{1}{\sin \alpha }=\frac{\sqrt{2}}{2\sin 2\alpha \cdot \cos \alpha }\\ or,\cos \alpha =\frac{1}{\sqrt{2}}\\ or,\alpha ={45}^0\\ \therefore \alpha =\beta ={45}^{0}and\gamma ={180}^0-2\left({45}^0\right)\\ hencewehave,\\ anglebetween\overrightarrow{A}and\overrightarrow{B}={180}^0-\gamma ={90}^0\\ anglebetween\overrightarrow{B}and\overrightarrow{C}={180}^0-\alpha ={135}^0\\ anglebetween\overrightarrow{C}and\overrightarrow{A}={180}^0-\beta ={135}^0\end{array}$

Hence,

option $\left(D\right)$ is correct answer.