Question

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

a). role="math" localid="1647759176820" $\overrightarrow{A}=3\hat{i}+4\hat{j}$and $\overrightarrow{B}=2\hat{i}-6\hat{j}$

b).$\overrightarrow{A}=3\hat{i}-2\hat{j}$and$\overrightarrow{B}=6\hat{i}+4\hat{j}$

Short answer

a). $\overrightarrow{A}·\overrightarrow{B}=-18$

b).$\overrightarrow{A}·\overrightarrow{B}=10$

Step-by-step solution

Step 1. Given information

We have,

$\overrightarrow{A}=3\hat{i}+4\hat{j}$and$\overrightarrow{B}=2\hat{i}-6\hat{j}$

also $\overrightarrow{A}=3\hat{i}-2\hat{j}$and$\overrightarrow{B}=6\hat{i}+4\hat{j}$

Step 2. Part a). dot product A→·B→ if A→=3i^+4j^ and B→=2i^-6j^

The dot product for these two vectors is in the form

$\overrightarrow{A}·\overrightarrow{B}=A_x{B}_x+A_y{B}_y$

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

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=A_x{B}_x+A_y{B}_y \\ =\left(3\hat{i}\right)\left(2\hat{i}\right)+\left(4\hat{j}\right)(-6\hat{j}) \\ =6{\hat{i}}^2-24{\hat{j}}^2 \end{gathered}$$

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

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=6{\hat{i}}^2-24{\hat{j}}^2 \\ =6-24 \\ =-18 \end{gathered}$$

Step 3. Part b). Dot product A→·B→ if A→=3i^-2j^ and B→=6i^+4j^

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

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=A_x{B}_x+A_y{B}_y \\ =\left(3\hat{i}\right)\left(6\hat{i}\right)+(-2\hat{j})\left(4\hat{j}\right) \\ =18{\hat{i}}^2-8{\hat{j}}^2 \\ =18-8 \\ =10 \end{gathered}$$