Algebraic description of Parallel vectors Let $a,b,c\in \Bbb R^3\setminus\{0\}$.
If I use the geometric meaning of cross product and inner product. i.e $a\times b=|a||b|\sin\theta\hat n$, $\langle a,b\rangle=|a||b|\cos\theta$, it is clear that $a,b$ is parallel if $\langle a,b\rangle =|a||b|$ or $a\times b=0$.
But if I take $a=\lambda b$ for some real $\lambda$ as the definition of $a,b$ being parallel, I am stuck at showing the equivalence between $\langle a,b\rangle =|a||b|$ and $a\times b=0$ and $a,b$ being parallel. i.e $\langle a,b\rangle =|a||b|$ iff $a\times b=0$ iff $a,b$ are parallel.
Any suggestion will be appreciated.
 A: Suppose that $a,b$ are non-zero vectors and $\langle a,b\rangle=|a||b|$. This means that the angle between the vectors is $0$ and that $a$ and $b$, which implies that they are linearly dependent, thus their cross product is zero.
Now suppose that $a,b$ are non-zero vectors with non-zero components, $m,n\neq 0$ are a real scalars, and  the cross product of $a$ and $b$ is $0$. Since we are in $\mathbb{R^3}$, let us define $a$ as $\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$ and $b$ as $\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.
Now, $$a\times b= (a_2b_3-a_3b_2)i-(a_1b_3-a_3b_1)j+(a_1b_2-a_2b_1)k=\vec{0}$$ so we must have that $a_2b_3=a_3b_2$ , $a_1b_3=a_3b_1$, and $a_1b_2=a_2b_1$, which also implies that $\frac{a_2}{a_3}=\frac{b_2}{b_3}$, $\frac{a_1}{a_3}=\frac{b_1}{b_3}$, and $\frac{a_1}{a_2}=\frac{b_1}{b_2}$. The only way that this is possible is if $b$ is linearly dependent on $a$ and vice versa.
Remark (probably not a profound one): This is extremely similar to similar triangles and proportional side lengths.
So suppose $b_1=ma_1$ and $b_3=ma_3$. Now suppose $b_2=na_2$. But $\frac{a_2}{a_3}=\frac{b_2}{b_3}=\frac{na_2}{ma_3}$ implies that $n=m$ so then $a=mb$. Now we have that $a$ is linearly dependent on $b$ so that the angle between the vectors is $0$ and thus the dot product is $|a||b|$.
WLOG, consider the case when $a_1=0$. Then we must have that some of $a_2,a_3,$ or $b_1$ be zero. Note that $a_2$ and $a_3$ can't both be zero, or else $a$ would be the zero vector.
Case 1: Let $a_2=0$. Then either $a_3$ or $b_2$ is zero. So $b_2$ is zero. Now for the cross product to be zero, we must have that $b_1$ be zero, so then vectors $a$ and $b$ each have their third components as free variables, which implies that they are linearly dependent. Since $a_2=0$ implies that $b_1=0$, having them both be zero at the same time would not yield anything new.
Case 2: Let $a_3=0$. Now, $a_2\neq 0$ so we must have that $b_3=0$. Then $b_1=0$. So now vectors $a$ and $b$ each have their second components as free variables and are linearly dependent.
Case 3: Let $b_1=0$. Then for $a\times b=0$, we must have that at least one of $a_2, b_3$ be zero and one of $a_3,b_2$ be zero. If we choose $b_3=0$, then $a_3=0$. If we choose $b_2=0$, then $a_2=0$. If we choose $a_2=0$, then $b_2=0$. Finally, if we choose $a_3=0$, then $b_3=0$. In each case, we get that $a$ and $b$ are linearly dependent.
So if even one component of $a$ or $b$ is zero, for the cross product to be zero, they must be linearly dependent.
Since you suggested using geometric definitions, $a\times b=0=|a||b|\sin{\theta}\Longrightarrow \theta =0$ for non-zero vectors.
The other implications are quite easy to prove IMO.
A: With $\vec a=\lambda \vec b$, the cross product tells us that:
$$\vec a\times \vec b= (\lambda b_2b_3-\lambda b_3b_2)\vec i-(\lambda b_1b_3-\lambda b_3b_1)\vec j+(\lambda b_1b_2-\lambda b_2b_1)\vec k=0$$
The dot product becomes:
$$\pmatrix{\lambda b_1\\\lambda b_2\\\lambda b_3}\cdot\pmatrix{b_1\\b_2\\b_3} = \lambda (b_1^2+ b_2^2+ b_3^2)$$
And $$|a|=|\lambda b|=\sqrt{\lambda^2 (b_1^2+ b_2^2+ b_3^2)}=\lambda\sqrt{(b_1^2+ b_2^2+ b_3^2)}$$
$$|b|=\sqrt{b_1^2+ b_2^2+ b_3^2}$$
so  $$|a||b|=\lambda (b_1^2+ b_2^2+ b_3^2)$$
