I am asked to solve the following:
Show that $\vec{u}$, $\vec{v}$ and $\vec{w}$ are coplanar if and only if one of them is a linear combination of the other two
The thing is: we are not supposed to use dot product.
My solution:
If
$$a\vec{u}+b\vec{v}+c\vec{w}=0$$
then the dot product by $\vec{u}$ gives
$$a \vec{u}\cdot \vec{u} + 0 + 0 = 0\\ a=0 $$ Dot product by $v$ gives $$0+b \vec{v} \cdot \vec{v} + 0=0\\ b=0$$
Dot product by $w$ gives
$$0+0+c \vec{w}\cdot \vec{w}=0\\ c=0$$
So one of them must be a linear combination of the other two (since only the trivial solution is possible).
How could I do this without using dot product?
Make sure you understand each step of the following argument.
Only if:
The vectors $\vec u, \vec v, \vec w$ are coplanar iff $\dim(\operatorname{span}(\vec u, \vec v, \vec w)) \le 2$. Because there are more than two vectors in the set $\{\vec u, \vec v, \vec w\}$, they must be linearly dependent. Thus we can write the vector $\vec 0$ as a nontrivial linear combination of $\vec u, \vec v, \vec w$; i.e.
$$a\vec u + b\vec v + c\vec w = \vec 0$$ has at least one other solution besides $a=b=c=0$.
WLOG, assume that $a\ne 0$. Then we can rearrange the above equation as $$\vec u = -\frac ba\vec v -\frac ca \vec w$$ which means that one of the vectors is a linear combination of the other two.
If:
WLOG, let $\vec u = \alpha \vec v + \beta \vec w$. Then $1\vec u-\alpha\vec v -\beta \vec w = \vec 0$. Therefore $\{\vec u, \vec v, \vec w\}$ is a linearly dependent set. Which means that $\dim(\operatorname{span}(\vec u, \vec v, \vec w)) \lt 3$, or equivalently $\dim(\operatorname{span}(\vec u, \vec v, \vec w)) \le 2$. This, by definition, means that $\vec u, \vec v, \vec w$ are coplanar.$\ \ \ \square$