The problem describes a right-angled triangle $ABC$ where $\angle B = 90^\circ$, and the lengths of sides $AB = 5$ and $AC = 12$. The altitude from point $A$ to side $BC$ is labeled as $AD$. The task is to compute the following dot products:

1. $\vec{AB} \cdot \vec{DC}$
2. $\vec{BD} \cdot \vec{CA}$
3. $\vec{AC} \cdot \vec{CD}$

First, let's clarify the geometry of the triangle:

• Since $\angle B = 90^\circ$, we can use the Pythagorean theorem to calculate the length of the hypotenuse $BC$: $BC = \sqrt{AB^2 + AC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

• Point $D$ is the foot of the altitude from $A$ to $BC$. In a right triangle, the altitude splits the hypotenuse into two segments, $BD$ and $DC$. These can be found using the geometric mean theorem: $BD = \frac{AB^2}{BC} = \frac{5^2}{13} = \frac{25}{13}, \quad DC = \frac{AC^2}{BC} = \frac{12^2}{13} = \frac{144}{13}$

Now, let's calculate the dot products:

1. $\vec{AB} \cdot \vec{DC}$:

   Since $AB$ and $DC$ are perpendicular (right angle at $B$), their dot product is: $\vec{AB} \cdot \vec{DC} = 0$

2. $\vec{BD} \cdot \vec{CA}$:

   Since $BD$ and $CA$ form an acute angle in triangle $ABC$, their dot product will depend on their magnitudes and the cosine of the angle between them. However, due to perpendicularity at point $A$, the dot product is again: $\vec{BD} \cdot \vec{CA} = 0$

3. $\vec{AC} \cdot \vec{CD}$:

   Similar to the first case, $AC$ and $CD$ are perpendicular to each other, hence: $\vec{AC} \cdot \vec{CD} = 0$

Final Answer:

• $\vec{AB} \cdot \vec{DC} = 0$
• $\vec{BD} \cdot \vec{CA} = 0$
• $\vec{AC} \cdot \vec{CD} = 0$

Let me know if you'd like further details on any of these steps!

Here are 5 related questions to expand on the topic:

1. How does the geometric mean theorem apply to a right-angled triangle?
2. What are the properties of dot products in Euclidean geometry?
3. Can you generalize the relationship between dot products and the angles between vectors?
4. How can you calculate the lengths of the segments $BD$ and $DC$ using another method?
5. How would the results change if the triangle was not right-angled?

Tip: When dealing with dot products, remember that perpendicular vectors always have a dot product of zero!