Question

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

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

Answer 1

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

$\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=0\Rightarrow A+\overrightarrow{B}=-\overrightarrow{C}$
$A^2+B^2+2ABcos{\theta }_1=C^2(\because A=B)$
$B^2+B^2+2ABcos{\theta }_1=2B^2(\because C=\sqrt{2}B)$
$2ABcos{\theta }_1=0\Rightarrow cos{\theta }_1=0\Rightarrow {\theta }_1={90}^{\circ }$
$\overrightarrow{B}+\overrightarrow{C}=-\overrightarrow{A}\Rightarrow B^2+C^2+2BCcos{\theta }_2=A^2$
$B^2+2B^2+2B\sqrt{2}Bcos{\theta }_2=B^2$
$2B^2(1+\sqrt{2}cos{\theta }_2)=-\frac{1}{\sqrt{2}}\Rightarrow {\theta }_2={135}^{\circ }$
Thus, angles are $({90}^{\circ },{135}^{\circ },{135}^{\circ })$