You normalize a vector distance by calculating the distance between two vectors that have each been independently normalized to unit length (length of 1). This ensures the distance is based on the vectors' direction, not their magnitude.
Understanding Vector Normalization and Distance
Here's a breakdown of vector normalization and its relationship to distance:
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Vector Normalization: A vector is normalized by dividing each of its components by its magnitude (length). The resulting vector has a magnitude of 1, pointing in the same direction as the original vector. Mathematically, if v is a vector, then its normalized form v̂ is calculated as:
v̂ = v / ||v||
where ||v|| represents the magnitude (Euclidean norm) of v.
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Vector Distance: The distance between two vectors typically refers to the Euclidean distance (straight-line distance). Given vectors a and b, the Euclidean distance
dis calculated as:d = ||a - b||
Normalizing the Distance
To normalize the distance between vectors, you first normalize each vector individually, and then calculate the distance between the normalized vectors.
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Normalize Vector a: Calculate â = a / ||a||
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Normalize Vector b: Calculate b̂ = b / ||b||
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Calculate the Normalized Distance: The normalized distance, d_normalized, is:
d_normalized = ||â - b̂||
Properties of Normalized Distance
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Range: The normalized Euclidean distance has a range between 0 and 2.
- 0: The vectors point in the same direction (identical normalized vectors).
- 2: The vectors point in opposite directions (normalized vectors are diametrically opposed).
- √2 (approximately 1.414): The vectors are orthogonal (perpendicular).
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Direction Focus: It measures the angular separation between the vectors. Larger distances imply larger angular differences.
Example
Let's say you have two vectors:
- a = [3, 4]
- b = [1, 0]
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Normalize a:
- ||a|| = √(3² + 4²) = 5
- â = [3/5, 4/5] = [0.6, 0.8]
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Normalize b:
- ||b|| = √(1² + 0²) = 1
- b̂ = [1/1, 0/1] = [1, 0]
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Calculate Normalized Distance:
- d_normalized = ||[0.6, 0.8] - [1, 0]|| = ||[-0.4, 0.8]|| = √((-0.4)² + (0.8)²) = √(0.16 + 0.64) = √0.8 ≈ 0.894
When to Use Normalized Distance
Normalized distance is useful when:
- You are interested in the direction of vectors, not their magnitudes.
- You want to compare vectors with different scales or units.
- You need a distance metric that is invariant to scaling.
- Cosine similarity is closely related to normalized distance, as cosine similarity can be calculated as 1 - (normalized distance squared / 2)
In essence, normalizing the vectors before calculating the distance removes the influence of their magnitudes, focusing purely on the angular difference between them.
