Q 43E - University Physics with Modern Physics

Question

For the vectors $\overrightarrow{\mathbf{A}}\mathbf{,}\overrightarrow{\mathbf{B}}$ and $\overrightarrow{\mathbf{C}}$ in Fig. E1.24, find the scalar products $\left(a\right)\overrightarrow{\mathbf{A}}\mathbf{.}\overrightarrow{\mathbf{B}}\mathbf{;}\left(b\right)\overrightarrow{\mathbf{B}}\mathbf{.}\overrightarrow{\mathbf{C}}\mathbf{;}\left(a\right)\overrightarrow{\mathbf{C}}\mathbf{.}\overrightarrow{\mathbf{A}}\mathbf{.}$

Step-by-Step Solution

Answer

The scalar product of $\overrightarrow{A}.\overrightarrow{B},\overrightarrow{B}.\overrightarrow{C}$ and $\overrightarrow{A}.\overrightarrow{C}$ are $-100m^2,150m^2$ and $71m^2$.

Step 1: Given Data

$\begin{array}{l}\overrightarrow{A}=\left(8.00m\right)\\ \overrightarrow{B}=(15.0m)\\ \overrightarrow{C}=(12.0m)\end{array}$

Step 2: The scalar product of A → . B →

The angle of  $\overrightarrow{A}.\overrightarrow{B},150°$     

  $$\begin{gathered} \overrightarrow{A}.\overrightarrow{B}=AB \cos \theta \\ =\left(8\times 15\right)\cos 150 \\ =83.91\approx -100m^2 \end{gathered}$$                       

Step 3: The scalar product of B → . C →

The angle of $\overrightarrow{B}.\overrightarrow{C},145°$

$$\begin{gathered} \overrightarrow{B}.\overrightarrow{C}=BC \cos \theta \\ =\left(15\times 12\right)\cos 145 \\ =159.09\approx 150m^2 \end{gathered}$$

Step 4: The scalar product of A → . C →

The angle of $\overrightarrow{A}.\overrightarrow{B}, 65°$

$$\begin{gathered} \overrightarrow{A}.\overrightarrow{C}=AC \cos \theta \\ =\left(8\times 12\right) \cos 65 \\ =-53.99\approx 71m^2 \end{gathered}$$