If vec{a}_1 is the component of vector vec{a} | vedclass

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(C) The component of $\vec{a}$ along $\vec{b}$ is given by $\vec{a}_1 = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$.The componen

Vedclass · Accepted Answer

$\frac{(\vec{a} \cdot \vec{b}) (\vec{b} 	imes \vec{a})}{|\vec{b}|^2}$

Vedclass · Answer

(C) The component of $\vec{a}$ along $\vec{b}$ is given by $\vec{a}_1 = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} ight) \vec{b}$.The component of $\vec{a}$ perpendicular to $\vec{b}$ is $\vec{a}_2 = \vec{a} - \vec{a}_1 = \vec{a} - \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} ight) \vec{b}$.Now,calculate the cross product $\vec{a}_1 	imes \vec{a}_2$:$\vec{a}_1 	imes \vec{a}_2 = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} ight) 	imes \left( \vec{a} - \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} ight)$Using the distributive property of the cross product:$= \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} (\vec{b} 	imes \vec{a}) - \frac{(\vec{a} \cdot \vec{b})^2}{|\vec{b}|^4} (\vec{b} 	imes \vec{b})$Since $\vec{b} 	imes \vec{b} = \vec{0}$,the second term becomes zero:$= \frac{(\vec{a} \cdot \vec{b}) (\vec{b} 	imes \vec{a})}{|\vec{b}|^2}$.

If $\vec{a}_1$ is the component of vector $\vec{a}$ along the direction of vector $\vec{b}$,and $\vec{a}_2$ is the component of $\vec{a}$ perpendicular to $\vec{b}$,then $\vec{a}_1 \times \vec{a}_2 = \dots$

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