I'm new to this particular topic and have been working through these tutorial questions.
For part $c.)$ I need to show vectors $u$ and $v$ form an orthogonal basis and then use that basis to form an orthonormal basis by normalising each vector.
I've begun with using the dot product for $u$ and $v$, such that $1.2$ + $3.1$ = $2.3$ = $6$, which is not equal to 1. And from my understanding this needs to be equal to 1 to be considered orthonormal?
What am I doing wrong or missing here?
To be orthogonal means $\langle u,v\rangle = 0$. In this case $\langle u,v\rangle=2(2 \cdot 1)-2\cdot 3-1 \cdot 1 + 1 \cdot 3=0$ so the are orthogonal with respect to this inner product.
A unit vector is a vector $x$ such that the norm (ie. length) $||x|| = \sqrt{\langle x,x\rangle} = 1$. For a set ot vectors to be orthonormal means they are all orthogonal to each other and all unit vectors. Again using the inner product given we can calculate that $\langle u,u \rangle = 2(1 \cdot 1)-3 \cdot 1 - 1 \cdot 3 + 3 \cdot 3 = 5 \neq 1$ so it's not a unit vector.
We can however find a unit vector in the same direction as $u$ if we scale $u$ by the reciprocal of it's length, so $(1/||u||)u= u /||u|| = u / \sqrt{5} = (1/\sqrt{5},3/\sqrt{5})$ and we can see that this is a unit vector because inner products are bilinear, so in particular for some scalar $k$ then $\langle kx,y\rangle = \langle x,ky\rangle = k\langle x,y\rangle$. In this case $\langle u/\sqrt{5},u / \sqrt{5} \rangle = (1/\sqrt{5})^2 \langle u,u \rangle = (1/\sqrt{5})^2 5 = 5 / 5 = 1$ so $u / ||u||$ is a unit vector. You can calculate this directly from the inner product formula as well. Similiarly $v / ||v||$ will be a unit vector, making $u / ||u||$ and $v / ||v||$ orthonormal with respect to this inner product.