Q44E - University Physics with Modern Physics

Question

Question: Find the vector product $\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{B}}$ (expressed in unit vectors)

of the two vectors given in Exercise 1.38. What is the magnitude

of the vector product?

Step-by-Step Solution

Answer

The magnitude of $\left|\overrightarrow{\mathrm{A}}\right|=8.06,\left|\overrightarrow{\mathrm{B}}\right|=5.38$.

Step 1: Given data

$$\begin{gathered} \overrightarrow{\mathrm{A}}=4.00\mathrm{i} +7.00\mathrm{j} \\ \overrightarrow{\mathrm{B}}=5.00\mathrm{i} -2.00\mathrm{j} \end{gathered}$$

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

$$\begin{gathered} \overrightarrow{A}\times \overrightarrow{B}=\left[\begin{array}{ccc}\stackrel{⏜}{\mathrm{i}}& \stackrel{⏜}{j}& \stackrel{⏜}{k}\\ 4& 7& 0\\ 5& -2& 0\end{array}\right]\begin{array}{}\\ \end{array} \\ = \stackrel{⏜}{i} (0-0)-\stackrel{⏜}{j} (0-0)+\stackrel{⏜}{k} (-8-35) \\ =0\stackrel{⏜}{i }+0\stackrel{⏜}{j} -43\stackrel{⏜}{k } \\ =-43\stackrel{⏜}{k} \end{gathered}$$

Step 3: The magnitude of the vector product

The magnitude of vector is,

$$\begin{gathered} \left|\overrightarrow{\mathrm{A}}\right|=\sqrt{4^2+7^2}=\sqrt{65}=8.06 \\ \left|\overrightarrow{\mathrm{B}}\right|=\sqrt{5^2+2^2}=\sqrt{29}=5.3 \end{gathered}$$