Question

Evaluate the dot product $\overrightarrow{A}·\overrightarrow{B}$if

a. $\overrightarrow{A}=4\hat{i}-2\hat{j}$and $\overrightarrow{B}=-2\hat{i}-3\hat{j}$.

b. $\overrightarrow{A}=-4\hat{i}+2\hat{j}$and$\overrightarrow{B}=2\hat{i}+4\hat{j}$.

Short answer

a). $\overrightarrow{A}·\overrightarrow{B}=-2$

b).$\overrightarrow{A}·\overrightarrow{B}=0$

Step-by-step solution

Step 1. Given Information

We have $\overrightarrow{A}=4\hat{i}-2\hat{j}$and $\overrightarrow{B}=-2\hat{i}-3\hat{j}$

also$\overrightarrow{A}=-4\hat{i}+2\hat{j}$and$\overrightarrow{B}=2\hat{i}+4\hat{j}$

Step 2. Part a). Dot product A→·B→ if A→=4i^-2j^ and B→=-2i^-3j^

The dot product for these two vectors is given $\overrightarrow{A}·\overrightarrow{B}=A_x{B}_x+A_y{B}_y$in this form

Now, we put the $x-$components and $y-$components of the two vectors

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=A_x{B}_x+A_y{B}_y \\ =\left(4\hat{i}\right)(-2\hat{i})+(-2\hat{j})(-3\hat{j}) \\ =-8{\hat{i}}^2+6{\hat{j}}^2 \end{gathered}$$

The magnitudes of ${\hat{i}}^2=1$and ${\hat{j}}^2=1$. So, the dot product will be

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=-8{\hat{i}}^2+6{\hat{j}}^2 \\ =-8+6 \\ =-2 \end{gathered}$$

Step 3. Part b). Dot product A→·B→ if A→=-4i^+2j^and B→=2i^+4j^

We use the same step as in part (a) to get the dot product by putting the $x-$components and $y-$components of the two vector

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=A_x{B}_x+A_y{B}_y \\ =(-4\hat{i})\left(2\hat{i}\right)+\left(2\hat{j}\right)\left(4\hat{j}\right) \\ =-8{\hat{i}}^2+8{\hat{j}}^2 \\ =-8+8 \\ =0 \end{gathered}$$