Q46E - University Physics with Modern Physics

Question

For the two vectors in Fig. E1.35, find the magnitude and

direction of (a) the vector product $\overrightarrow{\mathbf{A}}\mathbf{ }\mathbf{x}\mathbf{ }\overrightarrow{\mathbf{B}}$; (b) the vector product $\overrightarrow{\mathbf{B}}\mathbf{ }\mathbf{x}\mathbf{ }\overrightarrow{\mathbf{A}}$.

Step-by-Step Solution

Answer

The direction of $\overrightarrow{\mathrm{B}} \mathrm{x} \overrightarrow{\mathrm{A}}$ is opposite to $\overrightarrow{\mathrm{A}} \mathrm{x} \overrightarrow{\mathrm{B}}$.

Step 1: Given data

               $$\begin{gathered} \left|\overrightarrow{\mathrm{A}}\right|=2.80\mathrm{cm} \\ \left|\overrightarrow{\mathrm{B}}\right|=1.90\mathrm{cm} \\ \mathrm{\theta }={60}^{\circ }+{60}^{\circ }={120}^{\circ } \end{gathered}$$

Step 2: The vector Product of A →   x   B →

$|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}} |=|\overrightarrow{\mathrm{A}} ||\overrightarrow{\mathrm{B}} | \mathrm{sin\theta }$

                $$\begin{gathered} =2.80\times 1.90\times \sin {120}^{\circ } \\ =2.80\times 1.90\times \sin ({90}^{\circ }+{30}^{\circ }) \\ =5.32\times \cos {30}^{\circ } \\ =5.32\times 0.8660 \\ =4.16\mathrm{cm} \end{gathered}$$

Step 3: The vector Product of

$|\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{A}} |=\left|\overrightarrow{\mathrm{B}}\right|\left|\overrightarrow{\mathrm{A}}\right| \mathrm{sin\theta }$

              $$\begin{gathered} =1.90\times 2.80\times \sin {120}^{\circ } \\ =4.61\mathrm{cm} \end{gathered}$$