Chapter 14.1, Problem 16WE

To determine

The magnitude and direction for the resultant vector formed by subtracting the two vectors.

Explanation of Solution

Given information:

The given vectors are,

  $\begin{array}{l}U:\text{ magnitude 218, bearing 22}°\text{.}\\ V:\text{ magnitude 170, bearing 112}°\text{.}\end{array}$

Calculations:

To get the distance and bearing of the resultant vector, first we draw the velocity diagram.

  

Here AB denotes the vector U. Its magnitude is $218$ and its direction is $22°$ .

Here BC denotes the vector V. Its magnitude is $170$ and its direction is opposite of $112°$ .

Therefore AC denotes the subtraction of U and V. From the diagram we can see that $\angle ABC=(112-22)°=90°$

Let ,

  $\angle BAC=\alpha$

Now to determine AC we have to apply cosine law.

Therefore,

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2-2\times AB\times BC\times \cos 90\\ A{C}^2={218}^2+{170}^2-2\times 218\times 170\times \cos 90\\ B{C}^2=76424\\ BC=\sqrt{764244}=276.44\end{array}$

Now to find the course of bearing we have to use law of sine.

  $\begin{array}{l}\frac{\sin \alpha }{170}=\frac{\sin 90}{276.44}\\ \sin \alpha =\frac{\sin 90}{276.44}\times 170\\ \sin \alpha =0.6149\\ \alpha ={\sin }^{-1}0.6149\\ \alpha =37.947°\end{array}$

Thus the magnitude of the resultant vector is $276.44$ . The course of bearing is $37.947°$