The component of vector overrightarrow{A}=a_x | vedclass

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(C) Let $\overrightarrow{B} = \hat{i} - \hat{j}$.To find the component of vector $\overrightarrow{A}$ along the direction of $\overrightarrow{B}$,we c

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$(a_x-a_y) / \sqrt{2}$

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(C) Let $\overrightarrow{B} = \hat{i} - \hat{j}$.To find the component of vector $\overrightarrow{A}$ along the direction of $\overrightarrow{B}$,we calculate the scalar projection of $\overrightarrow{A}$ onto the unit vector of $\overrightarrow{B}$.The unit vector along $\overrightarrow{B}$ is $\hat{b} = \frac{\overrightarrow{B}}{|\overrightarrow{B}|} = \frac{\hat{i} - \hat{j}}{\sqrt{1^2 + (-1)^2}} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$.The component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is given by $\overrightarrow{A} \cdot \hat{b}$.$\overrightarrow{A} \cdot \hat{b} = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \cdot \left( \frac{\hat{i} - \hat{j}}{\sqrt{2}} \right)$.$= \frac{a_x(1) + a_y(-1) + a_z(0)}{\sqrt{2}} = \frac{a_x - a_y}{\sqrt{2}}$.

The component of vector $\overrightarrow{A}=a_x \hat{i}+a_y \hat{j}+a_z \hat{k}$ along the direction of $\hat{i}-\hat{j}$ is

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