Modelling with vectors

In a nutshell

You are learning about vectors to solve problems in context. Vectors have both magnitude and direction. When modelling with vectors, it is important to note your positive and negative directions or, sign convention. In mechanics, the vector quantities have corresponding scalar quantities. The magnitude of a vector is its scalar quantity.

Sign convention

When modelling with vectors, it is important to note your positive and negative directions or, your sign convention. In mechanics, it is common to set East and North as the positive directions, using the unit vectors $$\textbf{i}$$ and $$\textbf{j}$$, respectively.

Example 1

Find the speed of a particle moving with the velocity $$(\textbf{i}-7\textbf{j})ms^{-1}$$ to $$3\ s.f.$$

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The speed is the scalar quantity, or magnitude, of the velocity.

$$speed = |\textbf{i}-7\textbf{j}|=\sqrt{1^2+(-7)^2}=\sqrt{50}=\underline{7.07ms^{-1}} (3\ s.f.)$$

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Note: Remember to include units.

Example 2

Find the distance moved by a particle which travels at velocity $$(-5\textbf{i}+\textbf{j})ms^{-1}$$ for $$1$$ minute to $$3\ s.f.$$

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You can use:

$$distance = speed \times time$$​​

Speed is the magnitude of velocity:

​$$speed = |-5\textbf{i}+\textbf{j}|=\sqrt{(-5)^2+1^2}=(\sqrt{26})ms^{-1}$$​​

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So:

​$$\begin{aligned}distance&=speed\times time\\&=(\sqrt{26})ms^{-1}\times 1min\\&=(\sqrt{26})ms^{-1}\times 60s\\&=\underline{306m}\ (3\ s.f.)\end{aligned}$$​​

Example 3

A hiker walks $$3km$$ due east from a fixed point $$O$$ to $$A$$, then $$2km$$ due south from $$A$$ to $$B$$. Find the position vector of $$B$$ relative to $$O$$, and the bearing of $$B$$ from $$O$$.

Sketch a diagram of the hiker's journey.

The position vector of $$B$$ can be written in terms of $$\textbf{i}$$ and $$\textbf{j}$$, with positive directions for East and North, respectively. Taking note of the sign convention. The hiker moves south, so in the $$-\textbf{j}$$ direction:

$$\underline{\overrightarrow{OB}= (3\textbf{i}-2\textbf{j})km}$$​​​

For the bearing of $$B$$ from $$O$$, you need the clockwise angle from North of $$O$$. This is $$90\degree+\theta$$. Using trigonometry:

$$\begin{aligned}\tan\theta&=\frac23\\\theta&=33.7\degree\end{aligned}$$​​

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A bearing is three figures and is always measured from north. So the bearing of $$B$$ from $$O$$:

$$90\degree+\theta=90\degree+33.7\degree=\underline{124\degree}$$​​