We are given an equilateral triangle with side length 2. $M$ is the midpoint of side $BC$. We need to find the dot products of the following vectors: (1) $\vec{AB} \cdot \vec{AC}$ (2) $\vec{CA} \cdot \vec{BC}$ (3) $\vec{AM} \cdot \vec{BC}$ (4) $\vec{BM} \cdot \vec{CM}$

1. Problem Description

We are given an equilateral triangle with side length

2. $M$ is the midpoint of side $BC$. We need to find the dot products of the following vectors:

(1) $\vec{AB} \cdot \vec{AC}$

(2) $\vec{CA} \cdot \vec{BC}$

(3) $\vec{AM} \cdot \vec{BC}$

(4) $\vec{BM} \cdot \vec{CM}$

2. Solution Steps

(1) $\vec{AB} \cdot \vec{AC}$

The formula for the dot product of two vectors is:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}$

In this case, $|\vec{AB}| = 2$, $|\vec{AC}| = 2$, and the angle between $\vec{AB}$ and $\vec{AC}$ is $60^{\circ}$. Therefore,

$\vec{AB} \cdot \vec{AC} = 2 \cdot 2 \cdot \cos{60^{\circ}} = 4 \cdot \frac{1}{2} = 2$

(2) $\vec{CA} \cdot \vec{BC}$

$|\vec{CA}| = 2$, $|\vec{BC}| = 2$, and the angle between $\vec{CA}$ and $\vec{BC}$ is $120^{\circ}$. Therefore,

$\vec{CA} \cdot \vec{BC} = 2 \cdot 2 \cdot \cos{120^{\circ}} = 4 \cdot (-\frac{1}{2}) = -2$

(3) $\vec{AM} \cdot \vec{BC}$

Since $M$ is the midpoint of $BC$, $AM$ is the altitude of the equilateral triangle. The length of the altitude is $\sqrt{2^2 - 1^2} = \sqrt{3}$. Thus $|\vec{AM}| = \sqrt{3}$ and $|\vec{BC}| = 2$. Also, $\vec{AM}$ is perpendicular to $\vec{BC}$, so the angle between them is $90^{\circ}$. Therefore,

$\vec{AM} \cdot \vec{BC} = \sqrt{3} \cdot 2 \cdot \cos{90^{\circ}} = \sqrt{3} \cdot 2 \cdot 0 = 0$

(4) $\vec{BM} \cdot \vec{CM}$

Since $M$ is the midpoint of $BC$, $|\vec{BM}| = 1$ and $|\vec{CM}| = 1$. The angle between $\vec{BM}$ and $\vec{CM}$ is $180^{\circ}$, since they point in opposite directions. Therefore,

$\vec{BM} \cdot \vec{CM} = 1 \cdot 1 \cdot \cos{180^{\circ}} = 1 \cdot (-1) = -1$

3. Final Answer

(1) $\vec{AB} \cdot \vec{AC} = 2$

(2) $\vec{CA} \cdot \vec{BC} = -2$

(3) $\vec{AM} \cdot \vec{BC} = 0$

(4) $\vec{BM} \cdot \vec{CM} = -1$