Question

For any two vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$, show that $\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}\mathbf{=}{\mathbf{A}}_{\mathbf{x}}{\mathbf{B}}_{\mathbf{x}}\mathbf{+}{\mathbf{A}}_{\mathbf{y}}{\mathbf{B}}_{\mathbf{y}}\mathbf{+}{\mathbf{A}}_{\mathbf{z}}{\mathbf{B}}_{\mathbf{z}}$ . Suggestions:Write $\overrightarrow{\mathbf{A}}$and $\overrightarrow{\mathbf{B}}$ in unit-vector form and use Equations 7.4 and 7.5.

Short answer

It is verified that$\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}={\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}{\mathrm{B}}_{\mathrm{y}}+{\mathrm{A}}_{\mathrm{z}}{\mathrm{B}}_{\mathrm{z}}$.

Step-by-step solution

Given data

There are two vectors $\overrightarrow{A}$ and$\overrightarrow{B}$.

Dot product between unit vectors

Equal unit vectors follow the following dot product rule

$$\begin{gathered} \hat{\mathbf{i}}\mathbf{·}\hat{\mathbf{i}}\mathbf{=}\mathbf{1} \\ \hat{\mathbf{j}}\mathbf{·}\hat{\mathbf{j}}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\left(I\right) \\ \hat{\mathbf{k}}\mathbf{·}\hat{\mathbf{k}}\mathbf{=}\mathbf{1} \end{gathered}$$

Unequal unit vectors follow the following dot product rule

$$\begin{gathered} \hat{\mathbf{i}}\mathbf{·}\hat{\mathbf{j}}\mathbf{=}\mathbf{0} \\ \hat{\mathbf{j}}\mathbf{·}\hat{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\left(\mathrm{II}\right) \\ \hat{\mathbf{k}}\mathbf{·}\hat{\mathbf{i}}\mathbf{=}\mathbf{0} \end{gathered}$$

Determining the dot product between two vectors

The two vectors $\overrightarrow{A}$and $\overrightarrow{B}$in unit vector form are

$$\begin{gathered} \overrightarrow{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k} \\ \overrightarrow{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k} \end{gathered}$$

Here $A_x,A_y,A_z,B_x,B_y,B_z$ are the components of the two vectors along the three axes.

The dot product in unit vector form is

$$\begin{gathered} \overrightarrow{A}.\stackrel{\rightharpoonup }{B}=A_x{B}_x\left(\hat{i}·\hat{i}\right)+A_x{B}_y\left(\hat{i}·\hat{j}\right)+A_x{B}_z\left(\hat{i}·\hat{k}\right) \\ +A_y{B}_x\left(\hat{j}·\hat{i}\right)+A_y{B}_y\left(\hat{j}·\hat{j}\right)+A_y{B}_z\left(\hat{j}·\hat{k}\right) \\ +A_z{B}_x\left(\hat{k}·\hat{i}\right)+A_z{B}_y\left(\hat{k}·\hat{j}\right)+A_z{B}_z\left(\hat{k}·\hat{k}\right) \end{gathered}$$

Use equations (I) and (II) to get

$$\begin{gathered} \overrightarrow{A}.\stackrel{\rightharpoonup }{B}=A_x{B}_x\left(1\right)+A_x{B}_y\left(0\right)+A_x{B}_z\left(0\right) \\ +A_y{B}_x\left(0\right)+A_y{B}_y\left(1\right)+A_y{B}_z\left(0\right) \\ +A_z{B}_x\left(0\right)+A_z{B}_y\left(0\right)+A_z{B}_z\left(1\right) \\ =A_x{B}_x+A_y{B}_y+A_z{B}_z \end{gathered}$$