Vectors

The idea

A vector has a size and a direction: position, velocity, force. Mechanics is full of them, and two ways of multiplying them do most of the work: the dot product (how much of one points along the other) and the cross product (turning effect, like a moment).

Writing vectors

• Components: $\mathbf{v}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$, where $\mathbf{i},\mathbf{j},\mathbf{k}$ are unit vectors along x, y, z.
• Magnitude: $|\mathbf{v}|=\sqrt{a^2+b^2+c^2}$
• Direction (2D): $\theta ={\tan }^{-1}(b/a)$. Check the quadrant.
• Unit vector: $\hat{\mathbf{v}}=\mathbf{v}/|\mathbf{v}|$, length 1.
• Add, subtract and scale component by component.

Dot product (gives a number)

$\mathbf{a}\cdot \mathbf{b}=a_x{b}_x+a_y{b}_y+a_z{b}_z=|\mathbf{a}||\mathbf{b}|\cos \theta$

• Use it to find the angle between two vectors.
• Perpendicular vectors have $\mathbf{a}\cdot \mathbf{b}=0$.
• The component of a along b is $\mathbf{a}\cdot \hat{\mathbf{b}}$. That’s how you resolve a force along a direction.

Cross product (gives a vector)

$\mathbf{a}\times \mathbf{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_x& a_y& a_z\\ b_x& b_y& b_z\end{array}\right||\mathbf{a}\times \mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin \theta$

• The result is perpendicular to both, with direction from the right-hand rule.
• Order matters: $\mathbf{a}\times \mathbf{b}=-\mathbf{b}\times \mathbf{a}$.
• Main use: the moment of a force, $\mathbf{M}=\mathbf{r}\times \mathbf{F}$, with $\mathbf{r}$ first.

Worked example

$\mathbf{a}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}$, $\mathbf{b}=\mathbf{i}-\mathbf{j}+4\mathbf{k}$. Find $\mathbf{a}\cdot \mathbf{b}$, the angle between them, and $\mathbf{a}\times \mathbf{b}$.

Answer

$\mathbf{a}\cdot \mathbf{b}=2-3-4=-5$. $|\mathbf{a}|=\sqrt{14}$, $|\mathbf{b}|=\sqrt{18}$. $\cos \theta =\frac{-5}{\sqrt{14}\sqrt{18}}=-0.315$ → $\theta =108.4°$ $\mathbf{a}\times \mathbf{b}=(3\cdot 4-(-1)(-1))\mathbf{i}-(2\cdot 4-(-1)\cdot 1)\mathbf{j}+(2(-1)-3\cdot 1)\mathbf{k}=11\mathbf{i}-9\mathbf{j}-5\mathbf{k}$ Check it’s perpendicular to $\mathbf{a}$: $22-27+5=0$ ✓

Common mistakes

• Forgetting the minus sign on the $\mathbf{j}$ term of the cross product.
• Swapping the order in $\mathbf{r}\times \mathbf{F}$, which flips the moment’s direction.
• Using degrees mode when the question is in radians, or the other way round.

Check yourself

1. A 50 N force acts in the +y direction at a point 0.3 m along x from a pivot. Moment?

$\mathbf{M}=0.3\mathbf{i}\times 50\mathbf{j}=15\mathbf{k}$ N m: 15 N m anticlockwise about z.

2. Are $3\mathbf{i}+4\mathbf{j}$ and $-4\mathbf{i}+3\mathbf{j}$ perpendicular?

Yes: $-12+12=0$.