Q48E - University Physics with Modern Physics

Question

For the two vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ in Fig. E1.39, find (a) the scalar product ; (b) the magnitude and direction of the vector product.

Step-by-Step Solution

Answer

The scalar product of  $\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=-6.62{\mathrm{m}}^2$and $\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=5.55{\mathrm{m}}^2\mathrm{k}^$.

Step 1: Given data

                  $$\begin{gathered} \overrightarrow{\mathrm{A}} =3.60\mathrm{m} \\ \overrightarrow{\mathrm{B}} =2.4\mathrm{m} \\ \mathrm{\theta }={90}^{\circ }+{20}^{\circ }+{30}^{\circ }={140}^{\circ } \end{gathered}$$       

Step 2: The scalar product

                 $$\begin{gathered} \overrightarrow{A}.\overrightarrow{\mathrm{B}}=\left|\overrightarrow{A}\right|\left|\overrightarrow{\mathrm{B}}\right|\cos \theta \\ =3.6\times 2.4\times \cos 140 \\ =-6.62m^2 \end{gathered}$$                 

Step 3: The scalar product A → × B →

                $$\begin{gathered} \overrightarrow{A}.\overrightarrow{\mathrm{B}}=\left|\overrightarrow{A}\right|\left|\overrightarrow{\mathrm{B}}\right|\sin \theta \hat{k} \\ =3.6\times 2.4\times \sin 140 \\ =5.55m^2\hat{k} \end{gathered}$$