91Ó°ÊÓ

The resultant vector of two or more vectors is the sum of their corresponding components along the axis.

Magnitude of vector ${\stackrel{→}{V}}_1$, $V_1=6.6\mathrm{units}$

Magnitude of vector ${\stackrel{→}{V}}_2$, $V_2=8.5\mathrm{units}$

Let the X-components of ${\stackrel{→}{V}}_1$and ${\stackrel{→}{V}}_2$be $V_{1\mathrm{x}}$and $V_{2\mathrm{x}}$, respectively; the Y-component be $V_{1\mathrm{y}}$and $V_{2\mathrm{y}}$, respectively, and the sum of both components be vector $\stackrel{→}{R}$.

As vector lies along the negative X-axis, its angle with the positive direction of X-axis is $180°$, and the angle with the positive Y-axis is $90°$.

Component of ${\stackrel{→}{V}}_1$along the X-axis is

$\begin{array}{c}{V}_{1\mathrm{x}}=6.6×\cos 180°\\ =-6.6\mathrm{units}.\end{array}$

Component of ${\stackrel{→}{V}}_1$along the Y-axis is

$\begin{array}{c}{V}_{1\mathrm{y}}=6.6×\cos 90°\\ =0\mathrm{units}.\end{array}$

Thus, the components of ${\stackrel{→}{V}}_1$along X and Y directions are –6.6 units and 0 units, respectively.

As the vector lies at the orientation of $+55°$from the positive X-axis, the angle it makes from the positive Y-axis is $\left(90°-55°\right)$.

Component of ${\stackrel{→}{V}}_2$along the X-axis is

$\begin{array}{c}{V}_{2\mathrm{x}}=8.5×\cos 55°\\ =4.88\mathrm{units}.\end{array}$

Component of ${\stackrel{→}{V}}_2$along the Y-axis is

$\begin{array}{c}{V}_{2\mathrm{y}}=8.5×\cos 35°\\ =6.96\mathrm{units}.\end{array}$

Thus, the components of ${\stackrel{→}{V}}_2$along X and Y directions are 4.88 units and 6.96 units, respectively.

Let the X-component of vector $\stackrel{→}{R}$be $R_{\mathrm{x}}$. Then, by the laws of vector algebra,

$\begin{array}{c}{R}_{\mathrm{x}}=V_{1\mathrm{x}}+V_{2\mathrm{x}}\\ =-6.6+4.88\\ =-1.72\mathrm{units}\end{array}$

Let the Y-component of vector $\stackrel{→}{R}$be $R_{\mathrm{y}}$. Then, by the laws of vector algebra,

$\begin{array}{c}{R}_{\mathrm{y}}=V_{1\mathrm{y}}+V_{2\mathrm{y}}\\ =0+6.96\\ =6.96\mathrm{units}\end{array}$

The magnitude of vector $\stackrel{→}{R}$can be written in terms of its components along the X and Y axes.

$\begin{array}{c}R=\sqrt{R_{\mathrm{x}}^2+R_{\mathrm{y}}^2}\\ =\sqrt{{\left(-1.72\right)}^2+{6.96}^2}\\ =7.17\mathrm{units}\end{array}$

Here, R is the magnitude of $\stackrel{→}{R}$.

Thus, the magnitude of $\stackrel{→}{R}$is 7.17 units.

From the values of X and Y components of $\stackrel{→}{R}$, the graph of $\stackrel{→}{R}$can be drawn as shown below.

Tangent of angle $\mathrm{θ}$made with the negative X-axis by vector $\stackrel{→}{R}$is

$\begin{array}{c}\tan \mathrm{θ}=\left|\frac{R_{\mathrm{y}}}{R_{\mathrm{x}}}\right|\\ =\frac{6.96}{1.72}\\ =4.05.\end{array}$

Taking ${\tan }^{-1}$both sides,

$\begin{array}{c}\mathrm{θ}={\tan }^{-1}\left(4.05\right)\\ =76.13°.\end{array}$

Thus, $\stackrel{→}{R}$makes a clockwise angle of $76.13°$with the negative X-axis.