Calculate the magnitude of the position vector $\mathbf{v} = (-3, 4)$

Solution

$|\mathbf{v}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Therefore the magnitude of $\mathbf{v}$ is $|\mathbf{v}| = 5$

What is the magnitude of the vector $\mathbf{U}$ if $\mathbf{U} = 4\mathbf{i} - 5\mathbf{j}$?

Solution

Given that $\mathbf{U} = 4\mathbf{i} - 5\mathbf{j}$

The vector $\mathbf{U} = 4\mathbf{i} - 5\mathbf{j}$ can be written as a position vector as $\mathbf{U} = (4, -5)$

$|\mathbf{u}| = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}$

So $|\mathbf{u}| = \sqrt{41}$

Definition: A unit vector is any vector whose magnitude or modulus is one.

Now if $\mathbf{U}$ is any vector, then the unit vector in the direction of $\mathbf{U}$ is given by $\frac{\mathbf{U}}{|\mathbf{U}|}$ and it is denoted by $\hat{U}$.

Find a unit vector in the direction of vector $\mathbf{U} = (12, 5)$

Solution

$\tilde{U} = \frac{U}{|U|}$

But $U = (12, 5)$ and $|U| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

Then

$\tilde{U} = \frac{U}{|U|} = \frac{(12, 5)}{13}$

$\tilde{U} = \left(\frac{12}{13}, \frac{5}{13}\right)$

To check that $\tilde{U} = \left(\frac{12}{13}, \frac{5}{13}\right)$ is a unit vector, you need to find its magnitude:

$|\tilde{U}| = \sqrt{\left(\frac{12}{13}\right)^2 + \left(\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169} + \frac{25}{169}} = \sqrt{\frac{169}{169}} = \sqrt{1} = 1$

The direction of a vector may be given by using either bearings or direction cosines.

(a) By Bearings

Bearings are angles from a fixed direction in order to locate the interested places on the earth's surface.

Reading bearings: There are two methods used to read bearings. In the first method all angles are measured with reference from the North direction only whereby the North is taken as 000°, the East is 090°, the South is 180° and West is 270°.

From the figure above, point P is located at a bearing of 050°, while Q is located at a bearing of 135°.

Commonly, the bearing of point B from point A is measured from the north direction at point A to the line joining AB, and that of A from B is measured from the north direction at point B to the line joining BA.

From the figure above, the bearing of B from A is 060° while that of A from B is 240°.

In the second method, two directions are used as reference directions: these are North and South.

In this method, the location of places is found by reading an acute angle from the north eastwards or westwards and from the south eastwards or westwards.

From the figure above, the direction of point A from O is N46°E, that of B is N50°W, while the direction of C is S20°E.

Mikumi is 140 km at a bearing of 070° from Iringa. Makambako is 160 km at a bearing of 215° from Iringa. Sketch the position of these towns relative to each other, hence calculate the magnitude and direction of the displacement from Makambako to Mikumi.

Solution

Let (1 cm) represent (20 km), and let (AB) be the displacement from Makambako to Mikumi. The following sketch describes the location of these two places relative to each other. Since (AB) is the displacement from Makambako to Mikumi, then the distance between (A) and (B) is the distance between the two places. This is to say point (A) stands for Makambako and (B) for Mikumi.

Sketch

Using the cosine rule:

$\begin{aligned} |\overrightarrow{AB}|^2 & = 160^2 + 140^2 - 2(160)(140) \cos 145^\circ \\ & = 25600 + 19600 - 44800 \cos 35^\circ \\ & = 45200 + 36700.16 \\ & = 81900.16 \\ \overrightarrow{AB} & = \sqrt{81900.16} \approx 286 \text{ km} \end{aligned}$

The displacement from Makambako to Mikumi is approximately 286 km.

By sine rule:

$$\begin{aligned} \frac{\sin A}{140^\circ} & = \frac{\sin 145^\circ}{286} \\ \sin A & = \frac{140 \times \sin 145^\circ}{286} \\ \sin A & = \frac{140 \times \sin 35^\circ}{286} = \frac{140 \times 0.5736}{286} \\ \sin A & = \frac{80.304}{286} = 0.280 \\ A & = 16.3^\circ \end{aligned}$$

The bearing of Mikumi from Makambako is $16.3^\circ + 35^\circ = 051.3^\circ$ or N51.3°W.

Alternatively, by using the scale AB is approximately 14.3 cm. Therefore AB = 14.3 × 20 km = 286 km and the bearing is obtained by protractor which is about N51°E.

(b) Direction Cosines

Where $\cos A$ and $\cos B$ are the direction cosines of OP.

Definition: If $\overrightarrow{OP} = (x, y)$ makes angles $\alpha$ and $\beta$ with the positive directions of x and y axes respectively, then the cosines of $\alpha$ and $\beta$ are the direction cosines of the vector $\overrightarrow{OP}$.

From the figure above, $P\hat{O}R = O\hat{P}Q = \beta$ and $P\hat{O}Q = O\hat{P}R = \alpha$, also $OP = |OP|$

Therefore:

$\cos \alpha = \frac{x}{|OP|} \quad \text{and} \quad \cos \beta = \frac{y}{|OP|}$

If $\mathbf{a} = 6\mathbf{i} + 8\mathbf{j}$ find the direction cosine of $\mathbf{a}$ and hence find the angle made by $\mathbf{a}$ with the positive x-axis.

Solution

Given $\mathbf{a} = 6\mathbf{i} + 8\mathbf{j}$

(a) $|\mathbf{a}| = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$

Let $\alpha$ and $\beta$ be the angles made by $\mathbf{a}$ with the positive x and y axes respectively.

Then:

$\cos \alpha = \frac{x}{|\mathbf{a}|} = \frac{6}{10} = \frac{3}{5}$

$\cos \beta = \frac{y}{|\mathbf{a}|} = \frac{8}{10} = \frac{4}{5}$

The direction cosines of $\mathbf{a}$ are $\frac{3}{5}$ and $\frac{4}{5}$.

To find $\alpha$, we consider the direction cosine of x-axis and read the angle corresponding to it from mathematical tables.

Now $\cos \alpha = \frac{3}{5}$, therefore $\alpha = 53.1^\circ$