Two vectors are given by $\mathbf{a}\mathbf{}\mathbf{⃗}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\hat{\mathbf{j}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{b}\mathbf{}\mathbf{⃗}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}$. Find localid="1656307855911" $\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{a}\mathbf{}\mathbf{⃗}\mathbf{\times }\mathbf{b}\mathbf{}\mathbf{⃗}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{a}\mathbf{}\mathbf{⃗}\mathbf{\cdot }\mathbf{b}\mathbf{}\mathbf{⃗}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{a}\mathbf{}\mathbf{⃗}\mathbf{+}\mathbf{b}\mathbf{}\mathbf{⃗}\mathbf{}\mathbf{)}\mathbf{\cdot }\mathbf{b}\mathbf{}\mathbf{⃗}\mathbf{,}\mathbf{}$and (d) the component of $\overrightarrow{\mathbf{a}}$along the direction of $\overrightarrow{\mathbf{b}}$.

This problem is based on the product rule in which the vector product and scalar product are the two ways of multiplying vectors. Also this problem involves the vector addition in which Components of all the vectors which are associated with the same unit can be added or subtracted.

The scalar product can be written as

$$\begin{gathered} \mathrm{a}⃗\cdot \mathrm{b}⃗=(\mathrm{a}{⃗}_{\mathrm{x}}\hat{\mathrm{i}}+\mathrm{a}{⃗}_{\mathrm{y}}\hat{\mathrm{j}}+\mathrm{a}{⃗}_{\mathrm{z}}\hat{\mathrm{k}})\cdot (\mathrm{b}{⃗}_{\mathrm{x}}\hat{\mathrm{j}}+\mathrm{b}{⃗}_{\mathrm{y}}\hat{\mathrm{j}}+\mathrm{b}{⃗}_{\mathrm{z}}\hat{\mathrm{k}}) \\ \mathrm{a}⃗\cdot \mathrm{b}⃗={\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{x}}+{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{y}}+{\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{z}}\left(\mathrm{i}\right) \end{gathered}$$

The vector product can be written as

$\mathrm{a}⃗\cdot \mathrm{b}⃗=(\mathrm{a}{⃗}_{\mathrm{x}}\hat{\mathrm{i}}+\mathrm{a}{⃗}_{\mathrm{y}}\hat{\mathrm{j}}+\mathrm{a}{⃗}_{\mathrm{z}}\hat{\mathrm{k}})\cdot (\mathrm{b}{⃗}_{\mathrm{x}}\hat{\mathrm{j}}+\mathrm{b}{⃗}_{\mathrm{y}}\hat{\mathrm{j}}+\mathrm{b}{⃗}_{\mathrm{z}}\hat{\mathrm{k}})\left(\mathrm{ii}\right)$

Given are,

$$\begin{gathered} \mathrm{a}⃗=3.0\hat{\mathrm{i}}+5.0\hat{\mathrm{j}} \\ \mathrm{b}⃗=2.0\hat{\mathrm{i}}+4.0\hat{\mathrm{j}} \end{gathered}$$

Here vector $\overrightarrow{\mathrm{a}}$and vector $\overrightarrow{\mathrm{b}}$does not have z component. Thus using equation (ii),$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$ can be written as

$$\begin{gathered} \mathrm{a}⃗\times \mathrm{b}⃗=({\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{y}}-{\mathrm{b}}_{\mathrm{x}}{\mathrm{a}}_{\mathrm{y}})\hat{\mathrm{k}} \\ \mathrm{a}⃗\times \mathrm{b}⃗=\left[\right(3\left)\right(4)-(5\left)\right(2\left)\right]\hat{\mathrm{k}}=2\hat{\mathrm{k}} \end{gathered}$$

Using equation (i), $\overrightarrow{\mathrm{a}}·\hat{\mathrm{b}}$can be written as

$$\begin{gathered} \mathrm{a}⃗\cdot \mathrm{b}⃗={\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{x}}+{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{y}} \\ \mathrm{a}⃗\cdot \mathrm{b}⃗=\left[\right(3\left)\right(2)+(5\left)\right(4\left)\right]=26 \end{gathered}$$

$$\begin{gathered} \mathrm{a}⃗+\mathrm{b}⃗=(3.0+2.0)\hat{\mathrm{i}}+(5.0+4.0)\hat{\mathrm{j}} \\ (\mathrm{a}⃗+\mathrm{b}⃗)\cdot \mathrm{b}⃗=(5.0)(2.0)+\left(9\right)\left(4\right) \\ (\mathrm{a}⃗+\mathrm{b}⃗)\cdot \mathrm{b}⃗=46 \end{gathered}$$

Consider

$$\begin{gathered} \hat{\mathrm{b}}=\frac{\mathrm{b}⃗}{|\mathrm{b}⃗|} \\ \hat{b}=\frac{2.0\hat{\mathrm{i}}+4.0\hat{\mathrm{j}}}{\sqrt{(2.0{)}^2+(4.0{)}^2}} \end{gathered}$$

The component of $\overrightarrow{\mathrm{a}}$along the direction of $\overrightarrow{\mathrm{b}}$can be written as

$$\begin{gathered} {\mathrm{a}}_{\mathrm{b}}=\mathrm{a}⃗\cdot \hat{\mathrm{b}}=\frac{\left(3\right)\left(2\right)+\left(5\right)\left(4\right)}{\sqrt{(2.0{)}^2+(4.0{)}^2}} \\ {\mathrm{a}}_{\mathrm{b}}=5.81 \end{gathered}$$