Working with vectors

​​In a nutshell

Vector quantities can be both positive or negative. Scalar quantities are always positive. 

Vector and scalar quantities

A vector quantity has both magnitude and direction, hence why it can be positive or negative. When an object is moving in the opposite direction to the positive direction, the vector quantity is negative. 

Vector quantities

Vectors can be written in the form $$x\ i\ +\ y\ j$$, where the $$i$$ value represents the horizontal component of the vector and the $$j$$ value represents the vertical component. Vectors can also be written as $$\begin{pmatrix}x\\y\\\end{pmatrix}$$, where the top value represents the $$i$$ component and the bottom value represents the $$j$$ component. When a vector is given using this notation, you can use Pythagoras' theorem to find its magnitude and trigonometry to find its direction.

Distance is the magnitude of displacement and speed is the magnitude of velocity. 

A scalar quantity has magnitude only, hence why it is always positive. 

Scalar quantities

Example 1

A particle is moving with a velocity of $$v=3i+4j$$. Find:

a: The speed of the particle.

b: The angle the direction of motion makes with the horixontal axis.

a: Speed is the magnitude of velocity:

$$|v| = \sqrt{3^2+4^2} = \sqrt{25} = 5$$

The speed of the particle is $$\underline{5\ ms^{-1}}.$$

​​

b: Use trigonometry to work out the angle. Components of velocity are the opposite and adjacent, so use tan:

​$$tan\theta= \dfrac43$$​​

​$$\theta=tan^{-1}\bigg(\dfrac43\bigg) = 53.13\degree$$​​

The direction of motion is $$\underline{53.1\degree}$$.

Example 2

A man walks from $$A$$ to $$B$$ and then from $$B$$ to $$C$$. His displacement from $$A$$ to $$B$$ is $$3i+5j$$. His displacement from $$B$$ to $$C$$ is $$6i-10j$$. Find: 

a: The distance from $$A$$ to $$C$$.

b: The total distance the man has travelled. 

a: $$\overrightarrow{AC} = \overrightarrow{AB} +\overrightarrow{BC}$$

​$$\overrightarrow{AC} = \begin{pmatrix}3\\5\end{pmatrix}+\begin{pmatrix}6\\-10\end{pmatrix} = \begin{pmatrix}9\\-5\end{pmatrix}$$​​

Distance is the magnitude of the displacement vector:

$$|\overrightarrow{AC}| = \sqrt{9^2+(-5^2)} = \sqrt{106} = 10.29563014$$​​

The distance from $$A$$ to $$C$$ is $$\underline{10.3\ m}$$.

b: The total distance travelled is $$|\overrightarrow{AB}| + |\overrightarrow{BC}|$$:

​$$|\overrightarrow{AB}|$$: 

​$$|\overrightarrow{AB}| = \sqrt{3^2+5^2} = \sqrt{34}$$

​$$|\overrightarrow{BC}|$$:​

​$$|\overrightarrow{BC}| = \sqrt{6^2+(-10^2)} = \sqrt{136}$$​​

Total distance travelled:

$$\sqrt{34} + \sqrt{136} = 17.4928557$$​​

​

The total distance travelled by the man is $$\underline{17.5\ m}$$.