Answer
The set $\{(3,-2),(-4,-6)\}$ is orthogonal, not orthonormal and it is a basis for $R^2$.
Work Step by Step
given $\{(3,-2),(-4,-6)\}$
let $v_1=(3,−2),v_2=(-4,-6)$
(a)
since $v_1v_2=-12+12=0$
then, the set $\{(3,-2),(-4,-6)\}$ is orthogonal.
(b)
scince
$\left\|{v}_{1}\right\|=\sqrt{v_{1} \cdot v_{1}}=\sqrt{9+4}==\sqrt{13}\neq1$
then, the set $\{(3,-2),(-4,-6)\}$ is not orthonormal.
(c)
by the corollary to Theorem 5.10, $\{(3,-2),(-4,-6)\}$ is a basis for $R^2$.
