(a) Write each vector in Fig. E1.39 in terms of the unit vectors $\widehat{\mathbf{i}}$ and $\widehat{\mathbf{j}}$. (b) Use unit vectors to express vector $\overrightarrow{\mathbf{C}}$, where $\overrightarrow{\mathbf{C}}\mathbf{=}\mathbf{3}\mathbf{.00}\overrightarrow{\mathbf{A}}\mathbf{-}\mathbf{4}\mathbf{.00}\overrightarrow{\mathbf{B}}$ (c) Find the magnitude and direction $\overrightarrow{\mathbf{C}}$.

The vector $\overrightarrow{C}=3\overrightarrow{A}-4\overrightarrow{B}$.

Consider a Vector quantity$\overrightarrow{R}=R_x\widehat{i}+R_y\widehat{j}$, Here$R_x,R_y$are the components along x,y, directions respectively and $\widehat{i},\widehat{j}$are the unit vectors along x,y directions respectively. The magnitude of this vector is expressed as,

$\left|\overrightarrow{R}\right|=\sqrt{{R_x}^2+{R_y}^2}$

Part (a)

$\overrightarrow{A}$ can be represented as $\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}$. Here A_X and A_y are the components of $\overrightarrow{A}$ in x and y directions respectively. $\overrightarrow{i}$and $\overrightarrow{j}$are the unit vectors in x and y directions respectively.

Using trigonometry, the components of $\overrightarrow{A}$ can be expressed as,

$\begin{array}{c}{A}_x=\left(3.60\text{\hspace{0.17em}m}\right)\cos \left(70°\right)\\ =1.23\text{\hspace{0.17em}m}\\ A_y=\left(3.60\text{\hspace{0.17em}m}\right)\sin \left(70°\right)\\ =3.38\text{\hspace{0.17em}m}\end{array}$

Thus, $\overrightarrow{A}$ can be represented as,

$\begin{array}{l}\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}\\ \overrightarrow{A}=\left(1.23\text{\hspace{0.17em}m}\right)\widehat{i}+\left(3.38\text{\hspace{0.17em}m}\right)\widehat{j}\end{array}$

Similarly, the components of $\overrightarrow{B}$ can be expressed as,

$\begin{array}{c}{B}_x=\left(-2.40\text{\hspace{0.17em}m}\right)\cos \left(30\right)\\ =-2.08\text{\hspace{0.17em}m}\\ B_{\text{y}}=\left(-2.40\text{\hspace{0.17em}m}\right)\sin \left(30\right)\\ =-1.2\text{\hspace{0.17em}m}\end{array}$

Thus, $\overrightarrow{B}$ can be represented as,

$\begin{array}{l}\overrightarrow{B}=B_x\widehat{i}+B_y\widehat{j}\\ \overrightarrow{B}=\left(-2.08\text{\hspace{0.17em}m}\right)\widehat{i}+\left(-1.20\text{\hspace{0.17em}m}\right)\widehat{j}\end{array}$

Thus, the representation of $\overrightarrow{A}$ and $\overrightarrow{B}$ in terms of unit vectors is $\overrightarrow{A}=\left(1.23\text{\hspace{0.17em}m}\right)\widehat{i}+\left(3.38\text{\hspace{0.17em}m}\right)\widehat{j}and\overrightarrow{B}=\left(-2.08\text{\hspace{0.17em}m}\right)\widehat{i}+\left(-1.20\text{\hspace{0.17em}m}\right)\widehat{j}$.

Part (b)

The vector $\overrightarrow{C}$ can be expressed as,

$\overrightarrow{C}=3\overrightarrow{A}-4\overrightarrow{B}$

Substitute $\left(1.23\text{\hspace{0.17em}m}\right)\widehat{i}+\left(3.38\text{\hspace{0.17em}m}\right)\widehat{j}for\overrightarrow{A}$$and\left(-2.08\text{\hspace{0.17em}m}\right)\widehat{i}+\left(-1.20\text{\hspace{0.17em}m}\right)\widehat{j}for\overrightarrow{B}$

$\begin{array}{c}\overrightarrow{C}=3\overrightarrow{A}-4\overrightarrow{B}\\ =3\left(\left(1.23\text{\hspace{0.17em}m}\right)\widehat{i}+\left(3.38\text{\hspace{0.17em}m}\right)\widehat{j}\right)-4\left(\left(-2.08\text{\hspace{0.17em}m}\right)\widehat{i}+\left(-1.20\text{\hspace{0.17em}m}\right)\widehat{j}\right)\\ =\left(3.69\text{\hspace{0.17em}m}\right)\widehat{i}+\left(10.14\text{\hspace{0.17em}m}\right)\widehat{j}-\left(-8.32\text{\hspace{0.17em}m}\right)\widehat{i}+\left(-4.8\text{\hspace{0.17em}m}\right)\widehat{j}\\ =\left(12.01\text{\hspace{0.17em}m}\right)\widehat{i}+\left(14.94\text{\hspace{0.17em}m}\right)\widehat{j}\end{array}$

The representation of $\overrightarrow{C}$in terms of unit vectors is $\overrightarrow{C}=\left(12.01\text{\hspace{0.17em}m}\right)\widehat{i}+\left(14.94\text{\hspace{0.17em}m}\right)\widehat{j}$.

Part (c)

The Vector $\overrightarrow{C}$ can be expressed as,

$\overrightarrow{C}=C_x\widehat{i}+C_y\widehat{j}$

Substitute 12.01 m for C_x and 14.94m for C_y,

$\overrightarrow{C}=\left(12.01\text{\hspace{0.17em}m}\right)\widehat{i}+\left(14.94\text{\hspace{0.17em}m}\right)\widehat{j}$

The magnitude of vector $\overrightarrow{C}$ can be expressed as,

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

Substitute 12.01 m for C_x , and 14.94 m for C_y ,

$\begin{array}{c}\left|\overrightarrow{C}\right|=\sqrt{{\left(12.01\text{\hspace{0.17em}m}\right)}^2+{\left(14.94\text{\hspace{0.17em}m}\right)}^2}\\ =\sqrt{367.4\text{\hspace{0.17em}m}}\\ =19.17\text{\hspace{0.17em}m}\end{array}$

The direction can be calculated as,

$\tan \theta =\frac{C_y}{C_x}$

Substitute 12.01 m for C_x and 14.94m for C_y .

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{14.94\text{\hspace{0.17em}m}}{12.01\text{\hspace{0.17em}m}}\right)\\ =51.2°\end{array}$

Thus, the magnitude of $\overrightarrow{C}$ is 19.17 m, and its directions is 51.2^o in the first quadrant.