91Ó°ÊÓ

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

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

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

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 $\stackrel{→}{A}$ can be represented as,

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

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

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

Substitute 4.00 for A_X , and 7.00 for A_Y.

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

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

$\begin{array}{l}\stackrel{→}{B}=5.00\widehat{i}−2.00\widehat{j}\\ B_x=5.00,\text{ }{B}_y=−2.00\end{array}$

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

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

Substitute 5.00 for B_xand -2.00 for B_y,

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

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

Part (b)

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

$\stackrel{→}{B}=5.00\widehat{i}−2.00\widehat{j}$

Consider vector $\stackrel{→}{C}$ is the resultant vector $\stackrel{→}{A}−\stackrel{→}{B}$of and is expressed as,$\stackrel{→}{C}=\stackrel{→}{A}−\stackrel{→}{B}$

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

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

Part (c)

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

$\begin{array}{l}\stackrel{→}{C}=−1.00\widehat{i}+9.00\widehat{j}\\ C_x=−1.00,\text{ }{C}_y=9.00\end{array}$

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

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

Substitute -1.00 for C_x and 9.00 for, C_y ,

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

Thus, the magnitude of $\stackrel{→}{A}−\stackrel{→}{B}$ is 9.06.

The direction of a vector quantity can be expressed as,

$\tan \mathrm{θ}=\frac{C_y}{C_x}$

Substitute -1.00 for C_x and 9.00 for, C_y ,

$\begin{array}{c}an\mathrm{θ}=\frac{9.00}{−1.00}\\ =−9.00\\ \mathrm{θ}={\tan }^{−1}\left(−9.00\right)+180°\\ =96.3°\end{array}$

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

Part (d)

The vector diagram of $\stackrel{→}{C}=\stackrel{→}{A}−\stackrel{→}{B}$ 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.