For the abstract scalar product or inner product see inner product space

In mathematics, the dot product, also known as the scalar product, is a binary operation which takes two vectors over the real numbers R and returns a real-valued scalar quantity. Thus it is the standard inner product of the Euclidean space.

The dot product of two vectors a = a₂, … , a_n and b = b₂, … , b_n is by definition

$\backslash mathbf\{a\}\backslash cdot\; \backslash mathbf\{b\}\; =\; a\_1b\_1\; +\; a\_2b\_2\; +\; \backslash cdots\; +\; a\_nb\_n\; =\; \backslash sum\_\{i=1\}^n\; a\_ib\_i$

where Σ denotes summation notation. For example, the dot product of two three-dimensional vectors 3, −2 and −2, −1 is

3, −2·−2, −1 = 1(4) + 3(−2) + −2(−1)= 0.

Using matrix multiplication and treating the (column) vectors as n×1 matrices, the dot product can also be written as:

$\backslash mathbf\{a\}\; \backslash cdot\; \backslash mathbf\{b\}\; =\; \backslash mathbf\{a\}^T\; \backslash mathbf\{b\}\; \backslash ,$

where a^T denotes the transpose of the matrix a. Using the example from above, this would result in a 1×3 matrix (i.e., vector) multiplied by a 3×1 vector (which, by virtue of the matrix multiplication, results in a 1×1 matrix, i.e., a scalar):

Geometric interpretation

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In the Euclidean space there is a strong relationship between the dot product and lengths and angles. For a vector a, a·a is the square of its length, and if b is another vector

$\backslash mathbf\{a\}\; \backslash cdot\; \backslash mathbf\{b\}\; =\; a\; \backslash ,\; b\; \backslash cos\; \backslash theta\; \backslash ,$

where a and b denote the length of a and b, and θ is the angle between them.

Since a·cos(θ) is the projection of a onto b, the dot product can be understood geometrically as the product of this projection with the length of b.

As the cosine of 90° is zero, the dot product of two perpendicular vectors is always zero. If a and b have length one (they are unit vectors), the dot product simply gives the cosine of the angle between them. Thus, given two vectors, the angle between them can be found by rearranging the above formula:

$\backslash theta\; =\; \backslash arccos\; \backslash left(\; \backslash frac\; \{\backslash bold\{a\}\backslash cdot\backslash bold\{b\}\}\; \{|\backslash bold\{a\}||\backslash bold\{b\}|\}\backslash right).$

Sometimes these properties are also used for defining the dot product, especially in 2 and 3 dimensions; this definition is equivalent to the above one. For higher dimensions the formula can be used to define the concept of angle.

The geometric properties rely on the basis of vectors being perpendicular and having unit length: either we start with such a basis, or we use an arbitrary basis and define length and angle (including perpendicularity) with the above.

As the geometric interpretation shows, the dot product is invariant under isometric changes of the basis: rotations, reflections, and combinations, keeping the origin fixed.

In other words, and more generally for any n, the dot product is invariant under a coordinate transformation based on an orthogonal matrix. This corresponds to the following two conditions:

• the new basis is again orthonormal (i.e. it is orthonormal expressed in the old one)
• the new base vectors have the same length as the old ones (i.e. unit length in terms of the old basis)

The dot product in physics

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In physics, for a vector a, a·a is the square of its magnitude, and if b is another vector

$\backslash mathbf\{a\}\; \backslash cdot\; \backslash mathbf\{b\}\; =\; a\; \backslash ,\; b\; \backslash cos\; \backslash theta\; \backslash ,$

where a and b denote the magnitude of a and b, and θ is the angle between them.

In physics, magnitude is a scalar in the physical sense, i.e. a physical quantity independent of the coordinate system, expressed as the product of a numerical value and a physical unit, not just a number. The dot product is also a scalar in this sense, given by the formula, independent of the coordinate system. The formula in terms of coordinates is evaluated with not just numbers, but numbers times units. Therefore, although it relies on the basis being orthonormal, it does not depend on scaling.

Example:

• Mechanical work is the dot product of force and displacement.

Properties

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The following properties hold if a, b, and c are vectors and r is a scalar.

The dot product is commutative:

$\backslash mathbf\{a\}\; \backslash cdot\; \backslash mathbf\{b\}\; =\; \backslash mathbf\{b\}\; \backslash cdot\; \backslash mathbf\{a\}.$

The dot product is bilinear:

$\backslash mathbf\{a\}\; \backslash cdot\; (r\backslash mathbf\{b\}\; +\; \backslash mathbf\{c\})$

= r(\mathbf{a} \cdot \mathbf{b}) +(\mathbf{a} \cdot \mathbf{c}).

The dot product is distributive:

$\backslash mathbf\{a\}\; \backslash cdot\; (\backslash mathbf\{b\}\; +\; \backslash mathbf\{c\})\; =\; \backslash mathbf\{a\}\; \backslash cdot\; \backslash mathbf\{b\}\; +\; \backslash mathbf\{a\}\; \backslash cdot\; \backslash mathbf\{c\}.$

When multiplied by a scalar value, dot product satisfies:

$(c\_1\backslash mathbf\{a\})\; \backslash cdot\; (c\_2\backslash mathbf\{b\})\; =\; (c\_1c\_2)\; (\backslash mathbf\{a\}\; \backslash cdot\; \backslash mathbf\{b\})$

(these last two properties follow from the first two).

Two non-zero vectors a and b are perpendicular if and only if a · b = 0.

If b is a unit vector, then the dot product gives the magnitude of the projection of a in the direction b, with a minus sign if the direction is opposite. Decomposing vectors is often useful for conveniently adding them, e.g. in the calculation of net force in mechanics.

Unlike multiplication of ordinary numbers, where if ab = ac, then b always equals c unless a is zero, the dot product does not obey the cancellation law:

• If a · b = a · c and a ≠ 0:
• Then we can write: a · (b - c) = 0 by the distributive law; and from the previous result above:
• If a is perpendicular to (b - c), we can have (b - c) ≠ 0 and therefore b ≠ c.

Generalization

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The inner product generalizes the dot product to abstract vector spaces, it is normally denoted by <a, b>. Due to the geometric interpretation of the dot product the norm ||a|| of a vector a in such an inner product space is defined as

$\backslash |\backslash mathbf\{a\}\backslash |\; =\; \backslash sqrt\{\backslash langle\backslash mathbf\{a\},\; \backslash mathbf\{a\}\backslash rangle\},$

such that it generalizes length, and the angle θ between two vectors a and b by

$\backslash cos\{\backslash theta\}\; =\; \backslash frac\{\backslash langle\backslash mathbf\{a\},\; \backslash mathbf\{b\}\backslash rangle\}\{\backslash |\backslash mathbf\{a\}\backslash |\; \backslash ,\; \backslash |\backslash mathbf\{b\}\backslash |\}.$

In particular, two vectors are considered orthogonal if their inner product is zero.

Proof of the geometric interpretation

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Note: This proof is shown for 3-dimensional vectors, but is readily extendable to n-dimensional vectors.

Consider a vector

$\backslash mathbf\{v\}\; =\; v\_1\; \backslash mathbf\{i\}\; +\; v\_2\; \backslash mathbf\{j\}\; +\; v\_3\; \backslash mathbf\{k\}.\; \backslash ,$

Repeated application of the Pythagorean theorem yields for its length v

$v^2\; =\; v\_1^2\; +\; v\_2^2\; +\; v\_3^2.\; \backslash ,$

But this is the same as

$\backslash mathbf\{v\}\; \backslash cdot\; \backslash mathbf\{v\}\; =\; v\_1^2\; +\; v\_2^2\; +\; v\_3^2,\; \backslash ,$

so we conclude that taking the dot product of a vector v with itself yields the squared length of the vector.

Lemma 1:$\backslash mathbf\{v\}\; \backslash cdot\; \backslash mathbf\{v\}\; =\; v^2.\; \backslash ,$

Now consider two vectors a and b extending from the origin, separated by an angle θ. A third vector c may be defined as

$\backslash mathbf\{c\}\; \backslash equiv\; \backslash mathbf\{a\}\; -\; \backslash mathbf\{b\}.\; \backslash ,$

creating a triangle with sides a, b, and c. According to the law of cosines, we have

$c^2\; =\; a^2\; +\; b^2\; -\; 2\; ab\; \backslash cos\; \backslash theta.\; \backslash ,$

Substituting dot products for the squared lengths according to Lemma 1, we get

$$

\mathbf{c} \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} - 2 ab \cos\theta. \,                   (1) But as c ≡ a − b, we also have

$$

\mathbf{c} \cdot \mathbf{c} = (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \,, which, according to the distributive law, expands to

$$

\mathbf{c} \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} -2(\mathbf{a} \cdot \mathbf{b}). \,                     (2) Merging the two c · c equations, (1) and (2), we obtain

$$

\mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} -2(\mathbf{a} \cdot \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} - 2 ab \cos\theta. \, Subtracting a · a + b · b from both sides and dividing by −2 leaves

$\backslash mathbf\{a\}\; \backslash cdot\; \backslash mathbf\{b\}\; =\; ab\; \backslash cos\backslash theta.\; \backslash ,$

Q.E.D.

See also

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• Cross product

External links

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• Java demonstration of dot product

Linear algebra | Binary operations

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