vec{a}, vec{b}, vec{c} are three unit vectors | vedclass

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(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Since $\vec{a} \perp \vec{b}$, we have $\vec{

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$-1$

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(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Since $\vec{a} \perp \vec{b}$, we have $\vec{a} \cdot \vec{b} = 0$. Given $|\vec{a}+\vec{b}+\vec{c}|=1$, squaring both sides gives $|\vec{a}+\vec{b}+\vec{c}|^2 = 1$. Expanding the dot product: $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = 1$. $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 1$. Substituting the known values: $1 + 1 + 1 + 2(0 + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 1$. $3 + 2(\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b}) = 1$. Since $\vec{c} \cdot \vec{a} = |\vec{c}||\vec{a}| \cos \alpha = \cos \alpha$ and $\vec{c} \cdot \vec{b} = |\vec{c}||\vec{b}| \cos \beta = \cos \beta$, we have: $3 + 2(\cos \alpha + \cos \beta) = 1$. $2(\cos \alpha + \cos \beta) = 1 - 3 = -2$. Therefore, $\cos \alpha + \cos \beta = -1$.

$\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $|\vec{a}+\vec{b}+\vec{c}|=1$ and $\vec{a}$ is perpendicular to $\vec{b}$. If $\vec{c}$ makes angles $\alpha, \beta$ with $\vec{a}, \vec{b}$ respectively, then $\cos \alpha+\cos \beta=$

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