91影视

The vector $\stackrel{鈫�}{A}$ is given as $\stackrel{鈫�}{A}=4.00\widehat{i}+7.00\widehat{j}$ and the vector $\stackrel{鈫�}{B}$ is given as, $\stackrel{鈫�}{B}=5.00\widehat{i}鈭�2.00\widehat{j}$.

The magnitude of vector $\stackrel{鈫�}{G}=G_x\widehat{i}+G_y\widehat{j}$ can be expressed as,

$\left|\stackrel{鈫�}{G}\right|=\sqrt{G_x^2+G_y^2}$

Here $G_x,G_y$are the components in x and y direction and $\widehat{i},\widehat{j}$ are the unit vectors inx and ydirections receptively.

Part (a)

The components of $\stackrel{鈫�}{A}$ can be represented as,

$\begin{array}{l}\stackrel{鈫�}{A}=4.00\widehat{i}+7.00\widehat{j}\\ A_x=4.00,\text{鈥�}{A}_y=7.00\end{array}$

The magnitude of $\stackrel{鈫�}{A}$ can be calculated as,

$\left|\stackrel{鈫�}{A}\right|=\sqrt{A_x^2+A_y^2}$

Substitute 4.00 for A_X , and 7.00 for A_Y.

$\begin{array}{c}\left|\stackrel{鈫�}{A}\right|=\sqrt{{\left(4.00\right)}^2+{\left(7.00\right)}^2}\\ =\sqrt{65}\\ =8.06\end{array}$

The components of $\stackrel{鈫�}{B}$can be represented as,

$\begin{array}{l}\stackrel{鈫�}{B}=5.00\widehat{i}鈭�2.00\widehat{j}\\ B_x=5.00,\text{鈥�}{B}_y=鈭�2.00\end{array}$

The magnitude of $\stackrel{鈫�}{B}$can be expressed as,

$\left|\stackrel{鈫�}{B}\right|=\sqrt{B_x^2+B_y^2}$

Substitute 5.00 for B_xand -2.00 for B_y,

$\begin{array}{c}\left|\stackrel{鈫�}{B}\right|=\sqrt{{\left(5.00\right)}^2+{\left(鈭�2.00\right)}^2}\\ =\sqrt{29}\\ =5.39\end{array}$

Thus, the magnitude $\stackrel{鈫�}{A}$of is 8.06 and magnitude of $\stackrel{鈫�}{B}$is 5.39.

Part (b)

The vector $\stackrel{鈫�}{A}$is given as $\stackrel{鈫�}{A}=4.00\widehat{i}+7.00\widehat{j}$ and the vector $\stackrel{鈫�}{B}$is given as,

$\stackrel{鈫�}{B}=5.00\widehat{i}鈭�2.00\widehat{j}$

Consider vector $\stackrel{鈫�}{C}$ is the resultant vector $\stackrel{鈫�}{A}鈭�\stackrel{鈫�}{B}$of and is expressed as,$\stackrel{鈫�}{C}=\stackrel{鈫�}{A}鈭�\stackrel{鈫�}{B}$

$$\begin{gathered} Substitute4.00\widehat{i}+7.00\widehat{j}for\stackrel{鈫�}{A},and5.00\widehat{i}鈭�2.00\widehat{j}for\stackrel{鈫�}{B}. \\ \begin{array}{c}\stackrel{鈫�}{C}=\left(4.00\widehat{i}+7.00\widehat{j}\right)鈭�\left(5.00\widehat{i}鈭�2.00\widehat{j}\right)\\ =\left(4.00鈭�5.00\right)\widehat{i}+\left(7.00鈭�\left(鈭�2.00\right)\right)\widehat{j}\\ =鈭�1.00\widehat{i}+9.00\widehat{j}\end{array} \end{gathered}$$

Thus, the vector difference $\stackrel{鈫�}{A}鈭�\stackrel{鈫�}{B}$ can be expressed as, $\stackrel{鈫�}{C}=鈭�1.00\widehat{i}+9.00\widehat{j}$

Part (c)

The components of $\stackrel{鈫�}{C}$ can be represented as,

$\begin{array}{l}\stackrel{鈫�}{C}=鈭�1.00\widehat{i}+9.00\widehat{j}\\ C_x=鈭�1.00,\text{鈥�}{C}_y=9.00\end{array}$

Thus, the magnitude of $\stackrel{鈫�}{C}$can be expressed as,

$\left|\stackrel{鈫�}{C}\right|=\sqrt{C_x^2+C_y^2}$

Substitute -1.00 for C_x and 9.00 for, C_y ,

$\begin{array}{c}\left|\stackrel{鈫�}{C}\right|=\sqrt{{\left(鈭�1.00\right)}^2+{\left(9.00\right)}^2}\\ =\sqrt{82}\\ =9.06\end{array}$

Thus, the magnitude of $\stackrel{鈫�}{A}鈭�\stackrel{鈫�}{B}$ is 9.06.

The direction of a vector quantity can be expressed as,

$\tan 胃=\frac{C_y}{C_x}$

Substitute -1.00 for C_x and 9.00 for, C_y ,

$\begin{array}{c}an胃=\frac{9.00}{鈭�1.00}\\ =鈭�9.00\\ 胃={\tan }^{鈭�1}\left(鈭�9.00\right)+180掳\\ =96.3掳\end{array}$

Thus, the $\stackrel{鈫�}{A}鈭�\stackrel{鈫�}{B}$ makes an angle of 96.3^o with x-axis.

Part (d)

The vector diagram of $\stackrel{鈫�}{C}=\stackrel{鈫�}{A}鈭�\stackrel{鈫�}{B}$ is plotted above. Which agrees with the results obtained in part (c). C_x is negative and C_y is positive. Thus, the resultant vector lies in the second quadrant.