How do these cross products work I do not understand the properties of these cross product problems. 
For #1: $(\vec{i} \times \vec{j} ) \times \vec{k} = \vec{k} \times \vec{k} = 0$.  So I assume that any vector, like "$\vec{k}$" crossed with itself will equal 0?
Then I have no idea how to solve these problems:
Number 2: $\text{ }$$\vec{k}\times (\vec{i}  - 2\vec{j})$
3: $\text{ }$$(\vec{j}-\vec{k}) \times (\vec{k} - \vec{j})$
4: $\text{ }$$(\vec{i} + \vec{j} ) \times (\vec{i} - \vec{j})$
Note: all the "$\times$"s are cross products. 
 A: If you have a vector $\vec{a}$, the cross product $\vec{a}\times\vec{a}$ is always zero. Actually, if $\vec{a}\times\vec{b}=0$, $\vec{a}$ and $\vec{b}$ are multiples of each other.
Note that for $k\times(i - 2j)$, we have $k\times i-2k\times j=j+2i$. This is one property: $\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}$.
For the third, we have $(j-k)\times(k - j)=j\times(k - j)-k\times(k - j)$ and finally $j\times k - j\times j-k\times k +k\times j=i-0-0-i=0$
The fourth you can do the same way.
A: $\times:\mathbb{R}^3\times\mathbb{R}^3\mapsto\mathbb{R}^3$ is an antisymmetric bilinear form.  This means that
$$
\left(a\vec{x}+b\vec{y}\right)\times\vec{z}=a\left(\vec{x}\times\vec{z}\right)+b\left(\vec{y}\times\vec{z}\right)\tag{1}
$$
and
$$
\vec{y}\times\vec{x}=-\vec{x}\times\vec{y}\tag{2}
$$
It is not associative; that is, one cannot claim that $\left(\vec{x}\times\vec{y}\right)\times\vec{z}=\vec{x}\times\left(\vec{y}\times\vec{z}\right)$.
An immediate consequence of $(2)$ is that
$$
x\times x=0\tag{3}
$$
because $x\times x=-x\times x$.
A consequence of $(1)$ and $(2)$ is that
$$
\begin{align}
\vec{z}\times\left(a\vec{x}+b\vec{y}\right)
&=-\left(a\vec{x}+b\vec{y}\right)\times\vec{z}\\
&=-a\left(\vec{x}\times\vec{z}\right)-b\left(\vec{y}\times\vec{z}\right)\\
&=a\left(\vec{z}\times\vec{x}\right)+b\left(\vec{z}\times\vec{y}\right)\tag{4}
\end{align}
$$
On the canonical basis for $\mathbb{R}^3$, $\{i,j,k\}$, $\times$ is defined by $(1)$, $(2)$, $(3)$, and
$$
\begin{align}
i\times j&=k\\
j\times k&=i\\
k\times i&=j
\end{align}\tag{5}
$$
Hopefully, applying these properties to your problems should help.
A: Hint: 1) use the fact that $$i\times j = k = - j \times i,$$ $$j\times k = i = - k \times j,$$ $$k\times i = j = - i \times k.$$ 2) use distributive law.
