To prove that three given vectors form a triangle, the most fundamental method is to demonstrate that when placed end-to-end (head-to-tail), they form a closed loop. This means their vector sum must equal the zero vector. This condition, combined with others relating to their directions and magnitudes, confirms the formation of a triangle.
Let's explore the key conditions required, incorporating the insights from your references.
The Vector Sum Condition
The primary way to show that three vectors, say $\vec{a}$, $\vec{b}$, and $\vec{c}$, can form a closed triangle is by proving their sum is the zero vector:
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$
Imagine placing the vectors one after another, with the tail of the second vector starting at the head of the first, and the tail of the third vector starting at the head of the second. If the head of the third vector ends exactly at the tail of the first vector, they form a closed shape – a triangle.
Why this is key:
- Closure: The sum being zero mathematically represents this geometric closure.
- Reference 3 Connection: If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, then we can rearrange this to, for example, $\vec{a} + \vec{b} = -\vec{c}$. Taking the magnitude of both sides, we get $|\vec{a} + \vec{b}| = |-\vec{c}| = |\vec{c}|$. This directly aligns with Reference 3: The magnitude of the sum of two vectors is equal to the magnitude of the third.
Non-Collinearity: Ensuring a Non-Degenerate Triangle
For the vectors to form a proper triangle (not just a straight line segment back and forth), they must not all lie on the same line.
- Reference 1: This is captured by Reference 1: All three vectors should have different directions. While "different directions" is a bit loose (vectors can be parallel but in opposite directions), the core idea is that they must not be collinear. If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and they are collinear, they would form a degenerate triangle where two vectors are in one direction and the third is opposite, simply tracing a line segment. For a true triangle, at least two vectors must have directions that are not parallel.
The Triangle Inequality: A Consequence of Formation
If the three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ form a triangle (i.e., $\vec{a} + \vec{b} + \vec{c} = \vec{0}$) and are non-collinear, their magnitudes must satisfy the triangle inequality.
- Reference 2: This states The sum of the magnitude of two of the vectors must be greater than the magnitude of the third one.
- $|\vec{a}| + |\vec{b}| > |\vec{c}|$
- $|\vec{a}| + |\vec{c}| > |\vec{b}|$
- $|\vec{b}| + |\vec{c}| > |\vec{a}|$
How this relates: Since $\vec{a} + \vec{b} = -\vec{c}$, we know from the general vector triangle inequality that $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$. Substituting $|\vec{a} + \vec{b}| = |\vec{c}|$ (from the vector sum condition / Ref 3), we get $|\vec{c}| \le |\vec{a}| + |\vec{b}|$. The inequality is strict ($<$) because the vectors are non-collinear (Ref 1). If they were collinear, the equality ($=$) would hold, corresponding to a degenerate triangle.
Summary of Conditions
To prove three vectors form a non-degenerate triangle, you typically show:
| Condition | Description | Based on Reference |
|---|---|---|
| Vector Sum is Zero | $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ when placed head-to-tail. | Implies Ref 3 |
| Non-Collinearity | The vectors do not all lie on the same line (related to different directions). | Ref 1 |
| Magnitude Triangle Inequality | The sum of the magnitudes of any two vectors is greater than the third. | Ref 2 (Consequence) |
In essence: Prove that the vector sum is zero. This implies Reference 3. Then, verify they are non-collinear (Reference 1), which elevates Reference 2 (the magnitude inequality) from $\ge$ to $>$.
Practical Insight
When given three specific vectors (e.g., in component form like $\vec{a} = \langle 1, 2 \rangle$, $\vec{b} = \langle 3, -1 \rangle$, $\vec{c} = \langle -4, -1 \rangle$), the most straightforward proof method is:
- Calculate the vector sum: Find $\vec{a} + \vec{b} + \vec{c}$. If the result is $\langle 0, 0 \rangle$ (the zero vector), the closure condition is met.
- Check for collinearity: See if any vector is a scalar multiple of another. If $\vec{a} = k\vec{b}$ for some scalar $k$, they are collinear. If none are scalar multiples of each other, they are non-collinear.
- If the sum is zero and they are non-collinear, they form a triangle. The magnitude inequalities (Ref 2) will automatically hold.
Example:
Let $\vec{a} = \langle 2, 1 \rangle$, $\vec{b} = \langle -1, 3 \rangle$, $\vec{c} = \langle -1, -4 \rangle$.
- Vector Sum: $\vec{a} + \vec{b} + \vec{c} = \langle 2 + (-1) + (-1), 1 + 3 + (-4) \rangle = \langle 0, 0 \rangle$. The sum is zero.
- Collinearity: Is $\vec{a} = k\vec{b}$? $\langle 2, 1 \rangle = k \langle -1, 3 \rangle$. This would require $2 = -k$ and $1 = 3k$, giving $k=-2$ and $k=1/3$, which is impossible. Thus, $\vec{a}$ and $\vec{b}$ are not collinear. Similarly, check other pairs. Since the sum is zero and they are not all on the same line, they are non-collinear (Ref 1).
- Conclusion: Since the vector sum is zero and they are non-collinear, these three vectors form a triangle. As a result, $|\vec{a}| + |\vec{b}| > |\vec{c}|$, etc. (Ref 2) and $|\vec{a} + \vec{b}| = |\vec{c}|$ (Ref 3) will also be true.
In summary, proving three vectors form a triangle is primarily done by demonstrating that their sum is the zero vector, confirming geometric closure, and ensuring they are non-collinear to form a proper triangular shape.
