In $\triangle A B C$, $\overline { A B } = 6 , \overline { A C } = 5 , \overline { B C } = 4$. Let $D$ be the midpoint of $\overline { A B }$, and $P$ be the intersection of the angle bisector of $\angle A B C$ and $\overline { C D }$, as shown in the figure. Select the correct options. (1) $\overline { C P } = \frac { 3 } { 7 } \overline { C D }$ (2) $\overrightarrow { A P } = \frac { 3 } { 7 } \overrightarrow { A B } + \frac { 2 } { 7 } \overrightarrow { A C }$ (3) $\cos \angle B A C = \frac { 3 } { 4 }$ (4) The area of $\triangle A C P$ is $\frac { 15 } { 14 } \sqrt { 7 }$ (5) (Dot product) $\overrightarrow { A P } \cdot \overrightarrow { A C } = \frac { 120 } { 7 }$
In $\triangle A B C$, $\overline { A B } = 6 , \overline { A C } = 5 , \overline { B C } = 4$. Let $D$ be the midpoint of $\overline { A B }$, and $P$ be the intersection of the angle bisector of $\angle A B C$ and $\overline { C D }$, as shown in the figure. Select the correct options.\\
(1) $\overline { C P } = \frac { 3 } { 7 } \overline { C D }$\\
(2) $\overrightarrow { A P } = \frac { 3 } { 7 } \overrightarrow { A B } + \frac { 2 } { 7 } \overrightarrow { A C }$\\
(3) $\cos \angle B A C = \frac { 3 } { 4 }$\\
(4) The area of $\triangle A C P$ is $\frac { 15 } { 14 } \sqrt { 7 }$\\
(5) (Dot product) $\overrightarrow { A P } \cdot \overrightarrow { A C } = \frac { 120 } { 7 }$