If overrightarrow{a}=hat{i}+3 hat{j}+13 hat{k | vedclass

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(C) The component of vector $\vec{a}$ along $\vec{b}$ is given by $\vec{p} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$.First,c

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$-\hat{i}+7 \hat{j}+10 \hat{k}$

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(C) The component of vector $\vec{a}$ along $\vec{b}$ is given by $\vec{p} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$.First,calculate the dot product $\vec{a} \cdot \vec{b} = (1)(2) + (3)(-4) + (13)(3) = 2 - 12 + 39 = 29$.Next,calculate the magnitude squared $|\vec{b}|^2 = (2)^2 + (-4)^2 + (3)^2 = 4 + 16 + 9 = 29$.So,the component of $\vec{a}$ along $\vec{b}$ is $\vec{p} = \left( \frac{29}{29} \right) \vec{b} = \vec{b} = 2\hat{i} - 4\hat{j} + 3\hat{k}$.The component of $\vec{a}$ perpendicular to $\vec{b}$ is $\vec{x} = \vec{a} - \vec{p} = \vec{a} - \vec{b}$.$\vec{x} = (\hat{i} + 3\hat{j} + 13\hat{k}) - (2\hat{i} - 4\hat{j} + 3\hat{k}) = (1-2)\hat{i} + (3+4)\hat{j} + (13-3)\hat{k} = -\hat{i} + 7\hat{j} + 10\hat{k}$.

If $\overrightarrow{a}=\hat{i}+3 \hat{j}+13 \hat{k}$ and $\overrightarrow{b}=2 \hat{i}-4 \hat{j}+3 \hat{k}$ are two vectors,then the component vector of $\vec{a}$ perpendicular to $\vec{b}$ is

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