For the vectors shown in Fig. 3–35, determine (a)$\overrightarrow{\mathbf{B}}-\mathbf{3}\overrightarrow{\mathbf{A}}$and (b)$\mathbf{2}\overrightarrow{\mathbf{A}}-\mathbf{3}\overrightarrow{\mathbf{B}}\mathbf{+}\mathbf{2}\overrightarrow{\mathbf{C}}$.

FIGURE 3-35

A vector represents the magnitude and the direction of a physical quantity.

The vector can be represented as

Here, the length from C to D represents the magnitude of the vector. The arrow at point D represents the direction of the vector. The vector is denoted by$\overrightarrow{\mathbf{A}}$.

From the figure 3-35 given in the question, the magnitude $A_x$in the x-direction can be determined as

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

The magnitude $A_y$in the y-direction can be expressed as

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

From figure 3-35 given in the question, the magnitude $B_x$in the x-direction can be determined as

$\begin{array}{c}{B}_x=-26.5\cos 56°\\ =-14.82\end{array}$

The magnitude $B_y$in the y-direction can be expressed as

$\begin{array}{c}{B}_y=26.5\sin 56°\\ =21.97\end{array}$

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

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

Here, $\varphi$is the angle of inclination of vector C with the x-axis.

The magnitude $C_x$in the x-direction fromfigure 3-35can be evaluated as

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

The magnitude $C_y$in the –y-direction can be given as

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

The x-component of vector $\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)$can be expressed as

$\begin{array}{c}\left|{\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)}_x\right|=B_x-3A_x\\ =-14.82-3\times 38.85\\ =-131.37\end{array}$

The y-component of vector $\left(\overrightarrow{\mathbf{B}}-\mathbf{3}\overrightarrow{\mathbf{A}}\right)$can be expressed as

$\begin{array}{c}\left|{\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)}_y\right|=B_y-3A_y\\ =21.97-3\times 20.657\\ =-40.001\end{array}$

The magnitude of vector $\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)$can be calculated as

$\left|\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)\right|=\sqrt{{\left(\left|{\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)}_x\right|\right)}^2+{\left(\left|{\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)}_y\right|\right)}^2}$.

Substituting the values in the above equation,

$\begin{array}{c}\left|\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)\right|=\sqrt{{\left(-131.37\right)}^2+{\left(-40.001\right)}^2}\\ =137.325\end{array}$

Thus, the magnitude of vector $\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)$is $137.325$.

The direction of the vector can be expressed as

$\begin{array}{c}\tan \theta =\frac{B_y-3A_y}{B_x-3A_x}\\ \theta ={\tan }^{-1}\left\{\frac{B_y-3A_y}{B_x-3A_x}\right\}\end{array}$

Substituting the values in the above equation,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-40.001}{-131.37}\right)\\ =16.935°\end{array}$

As both components of the vector are negative, the vector lies in the third quadrant inclined with the negative x-axis at an angle of $16.935°$in the anticlockwise direction.

Thus, the angle of inclination of vector $\left(\overrightarrow{\mathbf{B}}-3\overrightarrow{\mathbf{A}}\right)$ is $16.935°$ below the negative x-axis in the anticlockwise direction.

The magnitude of the x-component of vector $\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}+2\overrightarrow{\mathbf{C}}\right)$can be expressed as

$\begin{array}{c}\left|{\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}+2\overrightarrow{\mathbf{C}}\right)}_x\right|=\left(2A_x-3B_x+2C_x\right)\\ =\left(2\times 38.85-3\times \left(-14.82\right)+2\times 0\right)\\ =122.16\end{array}$

The y-component of vector $\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}+2\overrightarrow{\mathbf{C}}\right)$can be expressed as

$\begin{array}{c}\left|{\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}+2\overrightarrow{\mathbf{C}}\right)}_y\right|=\left(2A_y-3B_y+2C_y\right)\\ =\left(2\times 20.657-3\times 21.97+2\times \left(-31\right)\right)\\ =-86.596\end{array}$

The magnitude of vector $\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}\mathbf{+}2\overrightarrow{\mathbf{C}}\right)$can be calculated as

$\left|\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}\mathbf{+}2\overrightarrow{\mathbf{C}}\right)\right|=\sqrt{{\left(\left|{\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}\mathbf{+}2\overrightarrow{\mathbf{C}}\right)}_x\right|\right)}^2+{\left(\left|{\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}\mathbf{+}2\overrightarrow{\mathbf{C}}\right)}_y\right|\right)}^2}$

Substituting the values in the above equation,

$\begin{array}{c}\left|\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}\mathbf{+}2\overrightarrow{\mathbf{C}}\right)\right|=\sqrt{{\left(122.16\right)}^2+{\left(-86.596\right)}^2}\\ =149.739\end{array}$

Thus, the magnitude of vector $\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}\mathbf{+}2\overrightarrow{\mathbf{C}}\right)$is 149.739.

The direction of the vector can be expressed as

$\begin{array}{c}\tan \theta =\frac{2A_y-3B_y+2C_y}{2A_x-3B_x+2C_x}\\ \theta ={\tan }^{-1}\left\{\frac{2A_y-3B_y+2C_y}{2A_x-3B_x+2C_x}\right\}\end{array}$

Substituting the values in the above equation,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-86.596}{122.16}\right)\\ =-35.332°\end{array}$

The y-component of the vector is negative, and the x-component is positive. Here, the negative sign of the angle indicates that the vector lies inclined with the positive x-axis in the fourth quadrant at an angle of $35.332°$below the +x-axis in the clockwise direction.

Thus, the angle of inclination of vector $\left(2\overrightarrow{\mathbf{A}}-3\overrightarrow{\mathbf{B}}\mathbf{+}2\overrightarrow{\mathbf{C}}\right)$ is $35.332°$below the +x-axis in the clockwise direction.