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If $\bar{a}, \bar{b}, \bar{c}$ are the three unit vectors such that $|\bar{a}+\bar{b}+\bar{c}|=1$ and $\bar{b}$ is perpendicular to $\bar{c}$. If $\bar{a}$ makes angles $\alpha, \beta$ with $\bar{b}$ and $\bar{c}$ respectively, then $\cos \alpha+\cos \beta$ has the value
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The correct answer is:
$-1$
$\begin{aligned} & |\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^2=|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2+|\overrightarrow{\mathrm{c}}|^2+2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+2 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+2 \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}} \\ & \Rightarrow 1^2=1^2+1^2+1^2+2 \\ & |\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \alpha+2|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos 90^{\circ}+2|\overrightarrow{\mathrm{c}}||\overrightarrow{\mathrm{a}}| \cos \beta \\ & \Rightarrow 1+3+2 \cos \alpha+2 \times 0+2 \cos \beta \\ & \Rightarrow \cos \alpha+\cos \beta=-1\end{aligned}$
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