Question

Suppose the vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ on a plane satisfy the condition that $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=|\overrightarrow{a}+\overrightarrow{b}|=1;\overrightarrow{c}$ is perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}.\overrightarrow{c}>0$, then

Answer 1

${|\overline{a}+\overline{c}|}^2={\left|\overline{a}\right|}^2+{\left|\overline{c}\right|}^2+2\overline{a}.\overline{c}$

$\because \overline{a}isperpendicularto\overline{c}.$

$\therefore \overline{a}-\overline{c}=0-\left(i\right)$

$⟹{|\overline{a}+\overline{c}|}^2={\left|\overline{a}\right|}^2+{\left|\overline{a}\right|}^2$

$⟹{|\overline{a}+\overline{b}|}^2={\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+2\overline{a}.\overline{b}$

$\because \left|\overline{a}\right|=\left|\overline{b}\right|=|\overline{a}+\overline{b}|=1$

$⟹1=1+1+2\overline{a}.\overline{b}$

$⟹\overline{a}.\overline{b}=\frac{-1}{2}$

$⟹2\overline{a}.\overline{b}=-1$

$\overline{a}.\overline{b}=\frac{-1}{2}$

$⟹\left|\overline{a}\right|.\left|\overline{b}\right|\cos \theta =\frac{-1}{2}$

$⟹\cos \theta =\frac{-1}{2}$

$⟹\theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

$Anglebetween\overline{a}and\overline{b}=\frac{2\pi }{3}$

$\overline{a}.\overline{c}=0-from\left(i\right)$

$⟹\left|\overline{a}\right|.\left|\overline{c}\right|\cos \alpha =0$

$⟹\cos \alpha =0implies\alpha =\frac{\pi }{2}$

$Anglebetween\overline{a}and\overline{c}=\frac{\pi }{2}$