If vec{a} = 4 hat{i} + 5 hat{j} - 3 hat{k} an | vedclass

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(A) Given vectors are $\vec{a} = 4 \hat{i} + 5 \hat{j} - 3 \hat{k}$ and $\vec{b} = 6 \hat{i} - 2 \hat{j} - 2 \hat{k}$.The magnitude of the component o

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$2$

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(A) Given vectors are $\vec{a} = 4 \hat{i} + 5 \hat{j} - 3 \hat{k}$ and $\vec{b} = 6 \hat{i} - 2 \hat{j} - 2 \hat{k}$.The magnitude of the component of $\vec{b}$ parallel to $\vec{a}$ is given by the formula $\frac{|\vec{b} \cdot \vec{a}|}{|\vec{a}|}$.First,calculate the dot product $\vec{b} \cdot \vec{a} = (6)(4) + (-2)(5) + (-2)(-3) = 24 - 10 + 6 = 20$.Next,calculate the magnitude of $\vec{a}$,which is $|\vec{a}| = \sqrt{4^2 + 5^2 + (-3)^2} = \sqrt{16 + 25 + 9} = \sqrt{50} = 5 \sqrt{2}$.Now,substitute these values into the formula:Magnitude $= \frac{|20|}{5 \sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2}$.

If $\vec{a} = 4 \hat{i} + 5 \hat{j} - 3 \hat{k}$ and $\vec{b} = 6 \hat{i} - 2 \hat{j} - 2 \hat{k}$ are two vectors,then the magnitude of the component of $\vec{b}$ parallel to $\vec{a}$ is: (in $\sqrt{2}$)

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