91Ó°ÊÓ

The resultant of the vector can be obtained by the summation of two or more different vectors.

The effect of this resultant vector is the same as the addition of different vectors. The resultant vector also has magnitude as well as direction.

The given data can be listed below as:

• The magnitude of the vector is $M=90\mathrm{units}$.
• The y-component of the vector is $y=-65\mathrm{units}$.
• The resultant vector in the –x-direction is $\stackrel{→}{\mathbf{R}}=-80\mathrm{units}\hat{\mathbf{i}}$.

The magnitude can be expressed as:

$\begin{array}{c}{M}^2=x^2+y^2\\ x^2=M^2-y^2\end{array}$

Here, x is the component along the x-direction.

Substitute the values in the above equation.

$\begin{array}{c}{x}^2={\left(90\mathrm{units}\right)}^2-{\left(-65\mathrm{units}\right)}^2\\ x=\sqrt{8100-4225}\mathrm{unit}\\ x=Â\pm 62.249\mathrm{unit}\end{array}$

Thus, the two possibilities of x are $62.249\mathrm{unit}$and $-62.249\mathrm{unit}$.

Consider the x component to be positive, whose value is 62.249 units. Then, one vector can be expressed as:

$\stackrel{→}{\mathbf{P}}=x\hat{\mathbf{i}}+y\hat{\mathbf{j}}$

Substitute the values as $62.249\mathrm{unit}$for x and $-65\mathrm{units}$for $y$in the above equation.

role="math" localid="1644315584698" $\stackrel{→}{\mathbf{P}}=\left(62.249\hat{\mathbf{i}}+\left(-65\right)\hat{\mathbf{j}}\right)\mathrm{units}……\left(I\right)$

The new vector can be expressed as:

$\stackrel{→}{\mathbf{Q}}=x_1\hat{\mathbf{i}}+y\hat{\mathbf{j}}$

Here, $x_1$is the new position along the x-axis. Consider the magnitude of y only.

Substitute the value $65\mathrm{units}$for $y$in the above equation.

role="math" localid="1644315714980" $\stackrel{→}{\mathbf{Q}}=\left(x_1\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}……\left(II\right)$

Now, add both vectors from equations (I) and (II).

$\stackrel{→}{\mathbf{R}}=\stackrel{→}{\mathbf{P}}+\stackrel{→}{\mathbf{Q}}$

Substitute the values as $\left(62.249\hat{\mathbf{i}}+\left(-65\right)\hat{\mathbf{j}}\right)\mathrm{units}$for $\stackrel{→}{\mathbf{P}}$and $\left(x_1\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}$for $\stackrel{→}{\mathbf{Q}}$, and $-80\mathrm{units}\hat{\mathbf{i}}$for $\stackrel{→}{\mathbf{R}}$in the above equation.

$\begin{array}{c}-80\mathrm{units}\hat{\mathbf{i}}=\left(62.249\hat{\mathbf{i}}+\left(-65\right)\hat{\mathbf{j}}\right)+\left(x_1\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}\\ -80\mathrm{units}\hat{\mathbf{i}}=\left(62.249+x_1\right)\hat{\mathbf{i}}\mathrm{units}\end{array}$

Now, write the magnitude of the vectors on both sides from the above equation.

$\begin{array}{c}-80\mathrm{units}=\left(62.249+x_1\right)\mathrm{units}\\ x_1=-142.249\mathrm{units}\end{array}$

Substitute the value in equation (II). Therefore, the new vector can be expressed as:

$\stackrel{→}{\mathbf{Q}}=\left(-142.249\mathrm{units}\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}$

Thus, the vector can be given as $\left(-142.249\mathrm{units}\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}$.

The magnitude of the vector$\left(-142.249\mathrm{units}\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}$ can be calculated as:

$\begin{array}{c}\left|\left(-142.249\mathrm{units}\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}\right|=\sqrt{{\left(-142.249\mathrm{units}\right)}^2+{\left(65\mathrm{units}\right)}^2}\\ =156.396\mathrm{units}\end{array}$

Thus, the magnitude of the vector$\left(-142.249\mathrm{units}\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}$is $156.396\mathrm{units}$.

The direction of the vector can be expressed as:

$\begin{array}{c}\tan \mathrm{θ}=\left\{\frac{65}{142.249}\right\}\\ =24.558°\end{array}$

Here, only the magnitude of the vector has been considered for the calculation of the inclination angle.

Since the x-components of the vector $\left(-142.249\mathrm{units}\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}$are negative, the vector lies in the second quadrant above the –x-axis at an angle of $24.558°$. The direction of the vector is $24.558°$north of west.

Thus, the angle of inclination of the vector $\left(-142.249\mathrm{units}\hat{\mathbf{i}}+65\hat{\mathbf{j}}\right)\mathrm{units}$ is $24.558°$above the –x-axis or above the west direction.