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Question

(A) Let $\vec{c} = \vec{a} + \vec{b}$.Given $\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$.Then $\vec{c} = (

Vedclass · Accepted Answer

$\frac{4}{\sqrt{29}} \hat{i}+\frac{3}{\sqrt{29}} \hat{j}-\frac{2}{\sqrt{29}} \hat{k}$

Vedclass · Answer

(A) Let $\vec{c} = \vec{a} + \vec{b}$.Given $\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$.Then $\vec{c} = (2+2)\hat{i} + (2+1)\hat{j} + (-5+3)\hat{k} = 4\hat{i} + 3\hat{j} - 2\hat{k}$.The magnitude of $\vec{c}$ is $|\vec{c}| = \sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$.The unit vector in the direction of $\vec{c}$ is given by $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.$\hat{c} = \frac{1}{\sqrt{29}}(4\hat{i} + 3\hat{j} - 2\hat{k}) = \frac{4}{\sqrt{29}}\hat{i} + \frac{3}{\sqrt{29}}\hat{j} - \frac{2}{\sqrt{29}}\hat{k}$.

Find the unit vector in the direction of the sum of the vectors,$\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k}$.

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