If vec{p}, vec{q}, text{ and } vec{r} are thr | vedclass

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(A) Let the magnitude of the vectors be $|\vec{p}| = |\vec{q}| = |\vec{r}| = a$. Since the vectors are mutually perpendicular,we have $\vec{p} \cdot \

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$\cos^{-1}(1/\sqrt{3})$

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(A) Let the magnitude of the vectors be $|\vec{p}| = |\vec{q}| = |\vec{r}| = a$. Since the vectors are mutually perpendicular,we have $\vec{p} \cdot \vec{q} = \vec{q} \cdot \vec{r} = \vec{r} \cdot \vec{p} = 0$. Let $\vec{v} = \vec{p} + \vec{q} + \vec{r}$. We want to find the angle $	heta$ between $\vec{p}$ and $\vec{v}$. The formula for the angle is $\cos 	heta = \frac{\vec{p} \cdot \vec{v}}{|\vec{p}| |\vec{v}|}$. First,calculate $\vec{p} \cdot \vec{v} = \vec{p} \cdot (\vec{p} + \vec{q} + \vec{r}) = \vec{p} \cdot \vec{p} + \vec{p} \cdot \vec{q} + \vec{p} \cdot \vec{r} = |\vec{p}|^2 + 0 + 0 = a^2$. Next,calculate $|\vec{v}|^2 = (\vec{p} + \vec{q} + \vec{r}) \cdot (\vec{p} + \vec{q} + \vec{r}) = |\vec{p}|^2 + |\vec{q}|^2 + |\vec{r}|^2 + 2(\vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{r} + \vec{r} \cdot \vec{p}) = a^2 + a^2 + a^2 + 0 = 3a^2$. So,$|\vec{v}| = \sqrt{3}a$. Now,$\cos 	heta = \frac{a^2}{a \cdot (\sqrt{3}a)} = \frac{1}{\sqrt{3}}$. Therefore,$	heta = \cos^{-1}(1/\sqrt{3})$.

If $\vec{p}, \vec{q}, \text{ and } \vec{r}$ are three mutually perpendicular vectors of the same magnitude,find the angle between $\vec{p}$ and $\vec{p} + \vec{q} + \vec{r}$.

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