Physics - Vectors Question with Solution | TestHub

Let,$\left|\overrightarrow{\text{A}}\right|=\left|\overrightarrow{\text{B}}\right|=\text{x}$
Then$\left|\overrightarrow{\text{C}}\right|=\sqrt{2}\text{x}$
Now$\overrightarrow{\text{A}}+\overrightarrow{\text{B}}=-\overrightarrow{\text{C}}$
$\left(\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\right)·\left(\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\right)=\left(-\overrightarrow{\text{C}}\right)·\left(-\overrightarrow{\text{C}}\right)$
$2\left(\text{A.B}\right)+{\text{A}}^2+{\text{B}}^2={\text{C}}^2$
$2\text{ABcos}\theta +{\text{x}}^2+{\text{x}}^2=2{\text{x}}^2$
$\text{cos}\theta =0$
$\theta =90^{\text{o}}$
$\Rightarrow$Angle between$\mathrm{A}$and$\mathrm{B}$is${90}^0$
Again$\overrightarrow{\text{A}}+\overrightarrow{\text{C}}=-\overrightarrow{\text{B}}$
$\left(\overrightarrow{\text{A}}+\overrightarrow{\text{C}}\right)·\left(\overrightarrow{\text{A}}+\overrightarrow{\text{C}}\right)=\left(-\overrightarrow{\text{B}}\right)\left(-\overrightarrow{\text{B}}\right)$
${\text{A}}^2+{\text{C}}^2+2\text{Acos}\theta ={\text{B}}^2$
$3{\text{x}}^2+2\text{.}\text{x}\sqrt{2}\text{x}\text{cos}\theta ={\text{x}}^2$
$\text{cos}\theta =\frac{-2}{2\sqrt{2}}=\frac{-1}{\sqrt{2}}$
$\theta =135^{\text{o}}$
i.e Angle between$\mathrm{A}$and$\mathrm{C}$is${135}^o$
Similarly, angle between$\mathrm{B}$and$\mathrm{C}$is${\theta }_2$
$\overrightarrow{\text{B}}+\overrightarrow{\text{C}}=-\overrightarrow{\text{A}}$
${\text{B}}^2+{\text{C}}^2+2\text{Bcos}\theta ={\text{A}}^2$
$\text{cos}\theta =\frac{-2}{2\sqrt{2}}=\left(\frac{-1}{\sqrt{2}}\right)$
${\theta }_2=135^{\text{o}}$