The triangle $ABC$ satisfies
$$\mathrm { AB } = 2 , \quad \mathrm { BC } = 3 , \quad \mathrm { CA } = 4 .$$
(1) When we set $\angle \mathrm { ABC } = \theta$, the inner product $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } }$ of the vectors $\overrightarrow { \mathrm { AB } }$ and $\overrightarrow { \mathrm { BC } }$ is
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \mathbf { AB } \cos \theta .$$
Finding the value of $\cos \theta$ from the law of cosines, we obtain
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \frac { \mathbf { C } } { \mathbf { D } } .$$
(2) We divide the side BC into $n$ equal parts by the points $\mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } , \cdots , \mathrm { P } _ { n - 1 }$ which are arranged in ascending order of the distance from B, and set $\mathrm { B } = \mathrm { P } _ { 0 } , \mathrm { C } = \mathrm { P } _ { n }$. We are to find the value of $\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } }$.
When we calculate the inner product of $\overrightarrow { \mathrm { AP } _ { k - 1 } }$ and $\overrightarrow { \mathrm { AP } _ { k } }$ using (1), we have
$$\overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \mathbf { E } + \frac { \mathbf { F } } { 2 n } + \mathbf { G }$$
Hence we obtain
$$\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \frac { \mathbf { IJ } } { \mathbf { K } }$$