91Ó°ÊÓ

$a=4, b=3 \mathrm{and} c=5.$

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,

$\stackrel{→}{\mathrm{a}}.\stackrel{→}{\mathrm{b}}=ab\cos \mathrm{θ}..............\left(1\right)$

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

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

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

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

We know that,

$\cos \left(180-\mathrm{θ}\right)=-\cos \left(\mathrm{θ}\right)$

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

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

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

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

$\cos \mathrm{φ}=\frac{3}{5}$

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

$$\begin{gathered} \stackrel{→}{b}.\stackrel{→}{c}=bc.c\cos \left(\mathrm{τ}\mathrm{τ}-\mathrm{φ}\right) \\ =-bc.\cos \left(\mathrm{θ}\right) \\ =-3×5×\left(\frac{3}{5}\right) \\ =-9\mathrm{units} \end{gathered}$$

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