3. Rolling a die twice yields the numbers $m, n$, and the vector $\boldsymbol{a}=(m, n)$ is formed. Then the probability that the angle between $\boldsymbol{a}$ and the vector $\boldsymbol{b}=$ $(1,-1)$ is an angle within a right triangle is ( ).
(A) $\frac{5}{12}$
(B) $\frac{1}{2}$
(C) $\frac{7}{12}$
(D) $\frac{3}{4}$
3. C.

Solution 1: Since $m, n$ can take $1,2, \cdots, 6$, the vector $\boldsymbol{a}$ has $6 \times 6=36$ possible positions. When $\boldsymbol{a}$ and $\boldsymbol{b}=(1,-1)$ form an angle that is an interior angle of a right triangle,
$$
\begin{array}{l}
\cos \langle\boldsymbol{a}, \boldsymbol{b}\rangle=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}||\boldsymbol{b}|}=\frac{m-n}{\sqrt{m^{2}+n^{2}} \cdot \sqrt{2}} \in[0,1) \\
\Leftrightarrow 0 \leqslant m-n<0 \text {. }
$$

The above inequality is obviously always true.
The number of solutions to the inequality $m-n \geqslant 0$ is
$$
6+5+4+3+2+1=21 \text {. }
$$

Therefore, the probability that the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ is an interior angle of a right triangle is $\frac{21}{36}=\frac{7}{12}$.
Solution 2: Let the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ be $\theta$.
When $m=n$, $\theta$ is a right angle, with 6 cases; for the remaining 30 cases, by symmetry, half have $m-n>0$, making $\theta$ an acute angle, and the other half have $m-n<0$, making $\theta$ an obtuse angle. Therefore, the number of cases where the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ is an interior angle of a right triangle is $6+15=21$, and the probability is $\frac{21}{36}=\frac{7}{12}$.