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Question

(A) Let $\vec{c}$ be the sum of vectors $\vec{a}$ and $\vec{b}$.$\vec{c} = \vec{a} + \vec{b} = (2\hat{i} - \hat{j} + \hat{k}) + (2\hat{j} + \hat{k})$$

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$\frac{1}{3}(2\hat{i} + \hat{j} + 2\hat{k})$

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(A) Let $\vec{c}$ be the sum of vectors $\vec{a}$ and $\vec{b}$.$\vec{c} = \vec{a} + \vec{b} = (2\hat{i} - \hat{j} + \hat{k}) + (2\hat{j} + \hat{k})$$\vec{c} = 2\hat{i} + (2 - 1)\hat{j} + (1 + 1)\hat{k} = 2\hat{i} + \hat{j} + 2\hat{k}$The magnitude of $\vec{c}$ is given by $|\vec{c}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.The unit vector in the direction of $\vec{c}$ is $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.$\hat{c} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} = \frac{1}{3}(2\hat{i} + \hat{j} + 2\hat{k})$.

Find the unit vector in the direction of the sum of vectors $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{j} + \hat{k}$.

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