Let $\hat{u}$ and $\hat{v}$ be two unit vectors
Compute the dot product between $\hat{u}-4\hat{v}$ and $\hat{u}+4\hat{v}$
How do I go about solving this? This is the first time I've encountered a question like this and I'm really confused. I would appreciate it if anyone is willing to explain.
If $\hat{u}$ and $\hat{v}$ are unit vectors, do I just multiply everything by $1$? Also, does $\hat{u} = \hat{i}$ and $\hat{v} = \hat{j}$ or am I thinking about this the wrong way?
Let $V$ be a vector space over the field $F$ with inner product $\langle \cdot \rangle$.
Remark
For $u\in V$ we define the norm of $u$ as $\mid \mid u \mid \mid =\sqrt{\langle u,u \rangle}$.
For $u,v\in V$ we define their inner product as $\langle u,v\rangle\in F$ and we know that the inner product is a bilinear form (it means that is linear in each coordinate).Therefore if we remember that a inner product is symetric,we can easily calculate the required inner product(assuming that $F=\mathbb{R}$ otherwise we should use $\langle u,v \rangle=\overline{\langle v,u \rangle}$ where $\overline{\langle v,u\rangle}$ denotes the usual complex conjugate of a complex number). $$\langle u-4v,u+4v \rangle=\langle u-4v,u \rangle+\langle u-4v,4v \rangle$$ $$=\langle u,u \rangle-4\langle v,u \rangle+4\langle u,v \rangle-16\langle v,v \rangle$$ $$=(\mid \mid u \mid \mid^2-4^2\mid \mid v\mid \mid^2)$$ Now if $u$ and $v$ are unitary vectors, what can you said about their norm?