Answer
a) $\mathcal{i\cdot j}=(1)(0)+(0)(1)+(0)(0)=0$
$\mathcal{j\cdot k}=(0)(0)+(1)(0)+(0)(1)=0$
$\mathcal{k\cdot i}=(0)(1)+(0)(0)+(1)(0)=0$
So $\mathcal{i\cdot j= j\cdot k = k\cdot i}=0$
b) $\mathcal{i\cdot i}=(1)(1)+(0)(0)+(0)(0)=1$
$\mathcal{j\cdot j}=(0)(0)+(1)(1)+(0)(0)=1$
$\mathcal{k\cdot k}=(0)(0)+(0)(0)+(1)(1)=1$
So $\mathcal{i\cdot i= j\cdot j = k\cdot k}=1$
Work Step by Step
We start by identifying the standard basis vectors $\mathcal{i, j}$ and $\mathcal{k}$ by its definition:
$$\mathcal{i} = \left \langle {1, 0, 0} \right \rangle$$ $$\mathcal{j} = \left \langle {0, 1, 0} \right \rangle$$ $$\mathcal{k} = \left \langle {0, 0, 1} \right \rangle$$
a) The statement: $\mathcal{i\cdot j= j\cdot k = k\cdot i}=0$
In the dot product we must multiply each component and add everything together, so we have
$\mathcal{i\cdot j}=(1)(0)+(0)(1)+(0)(0)=0$
$\mathcal{j\cdot k}=(0)(0)+(1)(0)+(0)(1)=0$
$\mathcal{k\cdot i}=(0)(1)+(0)(0)+(1)(0)=0$
So we prove that $\mathcal{i\cdot j= j\cdot k = k\cdot i}=0$
b) The statement: $\mathcal{i\cdot i= j\cdot j = k\cdot k}=1$
Following the same way as the previous paragraph, we have
$\mathcal{i\cdot i}=(1)(1)+(0)(0)+(0)(0)=1$
$\mathcal{j\cdot j}=(0)(0)+(1)(1)+(0)(0)=1$
$\mathcal{k\cdot k}=(0)(0)+(0)(0)+(1)(1)=1$
So we prove that $\mathcal{i\cdot i= j\cdot j = k\cdot k}=1$
