Answer
(a) the set is orthogonal.
(b) the set is orthonormal.
(c) the set is not a basis for $R^4$.
Work Step by Step
Let $u=(\frac{\sqrt 2}{2},0,0,\frac{\sqrt 2}{2})$ and $v=(0,\frac{\sqrt 2}{2},\frac{\sqrt 2}{2},0), w=(-\frac{1}{2},\frac{1}{2},-\frac{ 1}{2},\frac{1}{2})$, then we have
(a) since
$$u\cdot v=0, u\cdot w =-\frac{\sqrt{2}}{4}+\frac{\sqrt{2}}{4}= 0, v\cdot w =\frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{4}= 0$$
then the set is orthogonal.
(b) since $$\|u\|=\frac{1}{2}+\frac{1}{2}=1 , \|v\|=\frac{2}{4}+\frac{2}{4}=1, $$
$$\|w\|=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1, $$
then the set is orthonormal.
(c) sine $R^4$ has dimension $4$ and the set has three vectors, then it is not a basis for $R^4$.
