If the vector vec{a} = hat{i} + hat{j} + hat{ | vedclass

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(D) Let $\vec{v} = \hat{i} + \hat{j} + \hat{k}$. The magnitude is $|\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.The angle $\alpha$ between $\vec{v}$

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$\alpha = \beta = \gamma$

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(D) Let $\vec{v} = \hat{i} + \hat{j} + \hat{k}$. The magnitude is $|\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.The angle $\alpha$ between $\vec{v}$ and $\hat{i}$ is given by $\cos \alpha = \frac{\vec{v} \cdot \hat{i}}{|\vec{v}| |\hat{i}|} = \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i}}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$. Thus,$\alpha = \cos^{-1}(\frac{1}{\sqrt{3}})$.The angle $\beta$ between $\vec{v}$ and $\hat{j}$ is given by $\cos \beta = \frac{\vec{v} \cdot \hat{j}}{|\vec{v}| |\hat{j}|} = \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{j}}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$. Thus,$\beta = \cos^{-1}(\frac{1}{\sqrt{3}})$.The angle $\gamma$ between $\vec{v}$ and $\hat{k}$ is given by $\cos \gamma = \frac{\vec{v} \cdot \hat{k}}{|\vec{v}| |\hat{k}|} = \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{k}}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$. Thus,$\gamma = \cos^{-1}(\frac{1}{\sqrt{3}})$.Therefore,$\alpha = \beta = \gamma$.

If the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ makes angles $\alpha, \beta, \gamma$ with vectors $\hat{i}, \hat{j}, \hat{k}$ respectively,then:

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