Question

For the vectors in Fig. 3-32, with$\mathit{a}\mathbf{=}\mathbf{4}$,$\mathit{b}\mathbf{=}\mathbf{3}$, and$\mathit{c}\mathbf{=}\mathbf{5}$, calculate (a)$\overrightarrow{\mathbf{a}}\mathbf{.}\overrightarrow{\mathbf{b}}$, (b)$\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{c}}$, and (c)$\overrightarrow{\mathbf{b}}\mathbf{.}\overrightarrow{\mathbf{c}}$

Short answer

(a) The dot product $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}$, is 0 .

(b) The dot product $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}$is equal to -16 units.

(c) The dot product$\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}$ is equal to -9 units.

Step-by-step solution

Given data

$a=4,\hspace{0.33em}b=3\hspace{0.33em}\mathrm{and}\hspace{0.33em}c=5.$

Understanding the concept

Use the formula of scalar product of two vectors to find their products. To find the angle between the vectors, use the given diagram and trigonometry.

The formula for the dot product is,

$\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=ab\cos \theta ..............\left(1\right)$

(a) Calculate the scalar product a→.b→ 

From the figure, it is seen that the angle between $\overrightarrow{a}$and$\overrightarrow{b}$is$90°$ . Use equation (i) to calculate the dot product.

role="math" localid="1658469445125" $$\begin{gathered} \overrightarrow{a}.\overrightarrow{b}=ab\cos \left(90\right) \\ =0\mathrm{unit} \end{gathered}$$

As the $\cos \left(90\right)$is 0, the dot product of $\overrightarrow{a}$and$\overrightarrow{b}$ is equal to 0.

(b) Calculate the scalar product a→.c→

To find the angle between $\overrightarrow{a}$and $\overrightarrow{c}$, use the figure. From figure, it can be seen that the angle between$\overrightarrow{a}$and $\overrightarrow{c}$is $\left(180-\theta \right)$.

We know that,

$\cos \left(180-\theta \right)=-\cos \left(\theta \right)$

Now, use the value in equation (i) to find the dot product between $\overrightarrow{a}$and$\overrightarrow{c}$.

role="math" localid="1658469899162" $$\begin{gathered} \overrightarrow{a}.\overrightarrow{c}=ac\cos \left(180-\theta \right) \\ =-ac\cos \left(\theta \right) \\ =-4\times 5\times \left(\frac{4}{5}\right) \\ =-16\mathrm{units} \end{gathered}$$

Therefore, the dot product between $\overrightarrow{a}$and $\overrightarrow{c}$is -16 units.

(c) The scalar product of two vectors b→.c→

Use the concept used in part (b) to find the dot product of$\overrightarrow{b}$and$\overrightarrow{c}$. From figure,

$\cos \phi =\frac{3}{5}$

Now, use equation (i) to write the dot product between$\overrightarrow{b}$and $\overrightarrow{c}$.

$$\begin{gathered} \overrightarrow{b}.\overrightarrow{c}=bc.c\cos \left(\tau \tau -\phi \right) \\ =-bc.\cos \left(\theta \right) \\ =-3\times 5\times \left(\frac{3}{5}\right) \\ =-9\mathrm{units} \end{gathered}$$

Therefore, the dot product between $\overrightarrow{b}$and $\overrightarrow{c}$is -9 units .