Question 11.1: The Vector Product Two vectors lying in the xy plane are giv...
The Vector Product
Two vectors lying in the xy plane are given by the equations \overrightarrow{A}=2 \hat{i}+3 \hat{j} and \overrightarrow{B}=-\hat{i}+2 \hat{j}. Find \overrightarrow{A} \times \overrightarrow{B} and verify that \overrightarrow{A} \times \overrightarrow{B}=-\overrightarrow{B} \times \overrightarrow{A}.
Conceptualize Given the unit-vector notations of the vectors, think about the directions the vectors point in space. Draw them on graph paper and imagine the parallelogram shown in Figure 11.2 for these vectors.
Categorize Because we use the definition of the cross product discussed in this section, we categorize this example as a substitution problem.
Write the cross product of the two vectors:
\overrightarrow{A} \times \overrightarrow{B}=(2 \hat{i}+3 \hat{j}) \times(-\hat{i}+2 \hat{j})Perform the multiplication using the distributive law:
\overrightarrow{A} \times \overrightarrow{B}=2 \hat{i} \times(-\hat{i})+2 \hat{i} \times 2 \hat{j}+3 \hat{j} \times(-\hat{i})+3 \hat{j} \times 2 \hat{j}Use Equations 11.7a through 11.7d to evaluate the various terms:
\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0 (11.7a)
\hat{i} \times \hat{j}=-\hat{j} \times \hat{i}=\hat{k} (11.7b)
\hat{j} \times \hat{k}=-\hat{k} \times \hat{j}=\hat{i} (11.7c)
\hat{k} \times \hat{i}=-\hat{i} \times \hat{k}=\hat{j} (11.7d)
\overrightarrow{A} \times \overrightarrow{B}=0+4 \hat{k}+3 \hat{k}+0=7 \hat{k}To verify that \overrightarrow{A} \times \overrightarrow{B}=-\overrightarrow{B} \times \overrightarrow{A}, evaluate \overrightarrow{B} \times \overrightarrow{A}:
\overrightarrow{B} \times \overrightarrow{A}=(-\hat{i}+2 \hat{j})\times (2\hat{i}+3\hat{j})Perform the multiplication:
\overrightarrow{B} \times \overrightarrow{A}=(-\hat{i}) \times 2 \hat{i}+(-\hat{i}) \times 3 \hat{j}+2 \hat{j} \times 2 \hat{i}+2 \hat{j} \times 3 \hat{j}Use Equations 11.7a through 11.7d to evaluate the various terms:
\overrightarrow{B} \times \overrightarrow{A}=0-3 \hat{k}-4 \hat{k}+0=-7 \hat{k}Therefore, \overrightarrow{A} \times \overrightarrow{B}=-\overrightarrow{B} \times \overrightarrow{A}. As an alternative method for finding \overrightarrow{A} \times \overrightarrow{B}, you could use Equation 11.8. Try it!
\overrightarrow{A} \times \overrightarrow{B}=\left(A_y B_z-A_z B_y\right) \hat{i}+\left(A_z B_x-A_x B_z\right) \hat{j}+\left(A_x B_y-A_y B_x\right) \hat{k} (11.8)
