Answer
(a) the set is orthogonal
(b) the set is orthonormal
(c) the set is a basis for $R^3$.
Work Step by Step
Let $u=(\frac{\sqrt 2}{2},0,\frac{\sqrt 2}{2})$ and $v=(-\frac{\sqrt 6}{6},\frac{\sqrt 6}{3},\frac{\sqrt 6}{6}), w=(\frac{\sqrt 3}{3},\frac{\sqrt 3}{3},-\frac{\sqrt 3}{3})$, then we have
(a) since
$$u\cdot v=-\frac{\sqrt {12}}{12}+\frac{\sqrt{ 12}}{12}= 0$$
$$u\cdot w=-\frac{\sqrt {6}}{6}+\frac{\sqrt{ 6}}{6}= 0$$
$$v\cdot w=-\frac{\sqrt {18}}{18}+\frac{\sqrt{ 18}}{9}-\frac{\sqrt{ 18}}{18}= 0$$,
then the set is orthogonal.
(b) since $$\|u\|=\frac{2}{4}+\frac{2}{4}=1$$
$$\|v\|=\frac{6}{36}+\frac{6}{9}+\frac{6}{36}=1$$
$$\|w\|=\frac{3}{9}+\frac{3}{9}+\frac{3}{9}=1$$
the set is not orthogonal then it is not orthonormal.
(c) since the set is orthogonal, then it is a basis for$R^3$.
