#Maths Welcome to the second instalment in the vector series! I hope you've gotten a good idea of what vectors and scalars are at this point (and how they correlate with each other). We'll be taking a deep dive into operations that are for vectors and vectors only, listed below in terms of difficulty: # The Scalar (Dot Product) The easiest one to get used to - all you have to do is just multiply the vectors given linearly - make sure to match the $x$ coordinate with the corresponding $x$ coordinate, and so on. For example, if I had two vectors $a$ and $b$, we would express the dot product as follows: $a \cdot b = a_{x}b_{x} + a_{y}b_{y}$ Where the subscript $x$ and $y$ denotes the coordinate of each of the vectors. The general form of the dot product is given by: $|a \cdot b| = |a||b|\cos\theta$ As a result, we can find the angle between two vectors as shown: $\cos \theta = \frac{{|a \cdot b|}}{|a||b|}$ >[!Abstract]- Proving the Generalised Theorem for $\cos \theta$: >A diagram is in order to understand this proof: > >![[Pasted image 20230815220504.png]] >Again, we'll be using $a$ and $b$ as our arbitrary vectors. We can use the law of cosines to find the length (scalar) AB: >$AB^2 = |a|^2 + |b|^2 - 2|a||b|\cos c$ >The magnitude of $a$ can be broken down into: >$|a| = \sqrt{ a_{x}^2 + a_{y}^2 }$ >This applies to the vector $b$ as well. The line $AB$ can also be described as a component of $a$ and $b$: >$AB = b-a = \begin{vmatrix} b_{x} - a_{x} \\ b_{y} - a_{y} \\ \end{vmatrix}$ >Therefore squaring AB takes the magnitude: >$AB^2 = (b_{x} - a_{x})^2 + (b_{y} - a_{y})^2 = b_{x}^2 + b_{y}^2 + a_{x}^2 + a_{x}^2 -2b_{x}a_{x} - 2b_{y}a_{y}$ >We can therefore assemble the equation: >$b_{x}^2 + b_{y}^2 + a_{x}^2 + a_{y}^2 -2b_{x}a_{x} - 2b_{y}a_{y} = a_{x}^2 + a_{y}^2 + b_{x}^2 + b_{y}^2 - 2(\sqrt{ a_{x}^2 + a_{y}^2 })(\sqrt{ b_{x}^2 + b_{y}^2 })\cos \theta$ >Note that moving all the square terms over will allow them to cancel out. This yields: >$2a_{x}b_{x} +2a_{y}b_{y} = 2\sqrt{ (a_{x}^2 + a_{x}^2)(b_{x}^2 + b_{y}^2) }$ >Divide by two on both sides and rewrite the square root as $|a|$ and $|b|$: >$a_{x}b_{x} + {a_{y}}b_{y} = |a||b|\cos \theta$ >One last algebraic reshuffle thus proves the identity! >$\frac{{a_{x}b_{x} + a_{y}b_{y}}}{|a||b|} = \cos \theta$ ## Dot Product Identities # The Vector (Cross Product) Here's where it starts to get interesting. The general form of a cross product is given with a $\times$ symbol between the two variables. Often times, we The general formula for a cross product is given by: $|a \times b| = |a||b|\sin \theta$ Where $a$ and $b$ are both vectors respectively. >[!Success]- Cross Product Intuition >The presence of the variable $\sin \theta$ is indicative of a cross product value that decreases as the angle between the object decreases. This lets us extra >[!Abstract]- Proving the Cross Product ## Cross Product Identities There are many, many identities we have to use in cross products. If you checked the green box above, you'll notice that we've already proven one of them - that the cross product of a vector with itself will yield a value of 0. There are many, many more! See if you can unpack them yourself - but there will be descriptions provided if that is not the case. $k(a \times b) = ka \times b = a \times kb$ $a\times b = -b\times a$ $a\times(b\times c) = (a\times b) + (a\times c)$ <u style="font-size:18"><b>Proofs in Chronological Order:</b></u> >[!Abstract]- Vector Triple Product Proof