Find the unit vector in the direction of the | vedclass

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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} + 2 \hat{k}) + (-\hat{i} + \hat{j}

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$\frac{1}{\sqrt{26}}(\hat{i}+5 \hat{k})$

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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} + 2 \hat{k}) + (-\hat{i} + \hat{j} + 3 \hat{k})$$\vec{c} = (2-1) \hat{i} + (-1+1) \hat{j} + (2+3) \hat{k} = \hat{i} + 0 \hat{j} + 5 \hat{k} = \hat{i} + 5 \hat{k}$Now,calculate the magnitude of $\vec{c}$:$|\vec{c}| = \sqrt{1^2 + 0^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$The unit vector in the direction of $\vec{c}$ is given by $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.$\hat{c} = \frac{\hat{i} + 5 \hat{k}}{\sqrt{26}} = \frac{1}{\sqrt{26}} \hat{i} + \frac{5}{\sqrt{26}} \hat{k}$.

Find the unit vector in the direction of the sum of the vectors $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}+3 \hat{k}$.

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