91Ó°ÊÓ

A vector can be used to define the magnitude along with the direction of a physical quantity.

The resultant of two or more vectors can be calculated with the help of the addition or the subtraction of the horizontal and vertical components of the individual vectors.

The component of vector A in the x-direction can be calculated with the help offigure 3-35 given in the question as

$\begin{array}{c}{A}_x=44\cos 28°\\ =38.85\end{array}$

The component of vector A in the y-direction can be expressed as

$\begin{array}{c}{A}_y=44\sin 28°\\ =20.657\end{array}$

The inclination of the vector C from the x-axis in figure 3-35 can be given as

$\begin{array}{c}\mathrm{ϕ}=90°+90°+90°\\ =270°\end{array}$

Here, $\mathrm{Ï•}$is the angle of inclination of vector C with the positive x-axis in the clockwise direction.

The component of vector C in the x-direction can be calculated with the help offigure 3-35 given in the question as

$\begin{array}{c}{C}_x=31\cos \mathrm{ϕ}\\ =31\cos 270°\\ =0\end{array}$

The component of vector C in the –y-direction can be expressed as

$\begin{array}{c}{C}_y=31\sin \mathrm{ϕ}\\ =31\sin 270°\\ =-31\end{array}$

The magnitude of the component of vector $\left(\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}\right)$in the x-direction can be expressed as

$\begin{array}{c}\left|{\left(\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}\right)}_x\right|=A_x-C_x\\ =38.85-0\\ =38.85\end{array}$

The component of vector $\left(\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}\right)$in the y-direction can be expressed as

$\begin{array}{c}\left|{\left(\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}\right)}_y\right|=A_y-C_y\\ =20.657-\left(-31\right)\\ =51.657\end{array}$

The magnitude of the vector can be calculated as

$\left|\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}\right|=\sqrt{{\left(\left|{\left(\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}\right)}_x\right|\right)}^2+{\left(\left|{\left(\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}\right)}_y\right|\right)}^2}$.

Substituting the values in the above equation,

$\begin{array}{c}\left|\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}\right|=\sqrt{{\left(38.85\right)}^2+{\left(51.657\right)}^2}\\ =64.636\end{array}$.

Thus, the magnitude of the vector $\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}$is $64.636$.

The direction of the vector can be expressed as

$\begin{array}{c}\tan \mathrm{θ}=\frac{A_y-C_y}{A_x-C_x}\\ \mathrm{θ}={\tan }^{-1}\left\{\frac{A_y-C_y}{A_x-C_x}\right\}\end{array}$

Substituting the values in the above expression,

$\begin{array}{c}\mathrm{θ}={\tan }^{-1}\left(\frac{51.657}{38.85}\right)\\ =53.054°\end{array}$

Thus, the direction of the vector $\stackrel{→}{\mathbf{A}}-\stackrel{→}{\mathbf{C}}$is $53.054°$from the positive x-axis in the anticlockwise direction.