Answer
(a) the set is orthogonal.
(b) the set is orthonormal.
(c) the set is a basis for $R^4$.
Work Step by Step
Let $u=(\frac{\sqrt{10}}{10},0,0,\frac{3\sqrt{10}}{10})$ and $v=(0,0,1,0), w=(0,1,0,0), z=(-\frac{3\sqrt{10}}{10},0,0,\frac{\sqrt{10}}{10})$, then we have
(a) since
$$u\cdot v=0, \quad u\cdot w =-\frac{\sqrt{2}}{4}+\frac{\sqrt{2}}{4}= 0,\\
u\cdot z=-\frac{3}{10}+\frac{3}{10}=0, \quad v\cdot w = 0,\\
v\cdot z=0, \quad w\cdot z=0 $$
then the set is orthogonal.
(b) since $$\|u\|=\frac{10}{100}+\frac{90}{100}=1 , \quad \|v\|=1, $$
$$\|w\|=1, \quad \|z\|=\frac{90}{100}+ \frac{10}{100}=1$$
then the set is orthonormal.
(c) since the set is orthogonal and has four vectors, then it is not a basis for $R^4$.
