How to Normalize a Vector

Reviewed by Joseph Meyer

Last Updated: October 23, 2024 Fact Checked

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A vector is a geometric object that has direction and magnitude. It may be represented as a line segment with an initial point (starting point) on one end and an arrow on the other end, such that the length of the line segment is the magnitude of the vector and the arrow indicates the direction of the vector. Vector normalization is a common exercise in mathematics and it also has practical applications in computer graphics.

Part 1

Part 1 of 5:

Define Terms

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The unit vector of a vector A is the vector with the same initial point and direction as A, but with a length of 1 unit.^{[1]
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Research source} It can be mathematically proven that there is one and only one unit vector for each given vector A.
This is the process of identifying the unit vector for a given vector A.^{[2]
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A bound vector in Cartesian space has its initial point at the origin of the coordinate system, expressed as (0,0) in two dimensions. This allows you to identify a vector solely in terms of its terminal point.^{[3]
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By restricting ourselves to bound vectors, A = (x, y) where the coordinate pair (x,y) indicates the location of the terminal point for vector A.^{[4]
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Part 2

Part 2 of 5:

Analyze the Objective

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From the definition of the unit vector, we know that the initial point and direction of the unit vector is the same as the given vector A. Furthermore, we know the length of the unit vector is 1.^{[5]
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The only variable we need to calculate is the terminal point of the unit vector.

Part 3

Part 3 of 5:

Derive a Solution for the Unit Vector

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From the proportionality of similar triangles, you know that any vector that has the same direction as vector A will have a terminal point (x/c, y/c) for some c. Furthermore, you know the length of the unit vector is 1.^{[6]
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Research source} Therefore, by the Pythagorean Theorem, [x^2/c^2 + y^2/c^2]^(1/2) = 1 -> [(x^2 + y^2)/c^2]^(1/2) -> (x^2 + y^2)^(1/2)/c = 1 -> c = (x^2 + y^2)^(1/2). Therefore, the unit vector u for the vector A = (x, y) is given as u = (x/(x^2 + y^2)^(1/2), y/(x^2 + y^2)^(1/2))

Part 4

Part 4 of 5:

Normalize a Vector in 2 Dimensional Space

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Calculate the unit vector u = (x/(x^2 + y^2)^(1/2), y/(x^2 + y^2)^(1/2)) = (2/(2^2 + 3^2)^(1/2), 3/(2^2 + 3^2)^(1/2)) = (2/(13^(1/2)), 3/(13^(1/2))). Therefore, A = (2,3) normalizes to u = (2/(13^(1/2)), 3/(13^(1/2))).

Part 5

Part 5 of 5:

Normalize a Vector in n Dimensional Space

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Generalize the equation for vector normalization in space of any dimension.^{[7]
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Research source} A vector A (a, b, c, …), u = (a/z, b/z, c/z, …) where z = (a^2 + b^2 + c^2 …)^(1/2).

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What is the direction of cosine?

Donagan

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A cosine does not have direction. It is a ratio of one length to another.

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