Dot product

(Redirected from Inner product)

In mathematics, the dot product (also known as the scalar product and the inner product) is a sesquilinear function (·) : V × V → F, where V is a vector space over the field F, having some further properties. A formal definition is found in the article inner product space.

In other words, it maps a pair of vectors to a scalar (i.e. an element of the field). The inner product of a and b is sometimes denoted <a, b> or <a|b>.

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Scalarproduct.gif

It is intuitively defined as

[itex] \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \, |\mathbf{b}| \cos \theta \;[itex],

where |x| denotes the norm (or length) of x and θ is the angle between the two vectors.

Note: A formal definition of dot product is somewhat different from this; there, the angle between a and b is defined by the above equality.

Thus, the dot product of two perpendicular vectors is always zero. If a and b are both unit vectors (i.e., of length 1), 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:

[itex] \cos{\theta} = \frac{\mathbf{a \cdot b}}{|\mathbf{a}| \, |\mathbf{b}|} [itex]

This can be understood very easily: The first vector is projected onto the second vector (the order does not matter as the dot-product is commutative) by calculating the dot-product, and afterwards "normalized" by dividing the obtained scalar value of the numerator by their scalar lengths. Thus the scalar value of the fraction must be less than or equal to 1 and can be easily translated into a angular value (As the trigonometric functions are really nothing more than Taylor approximated functions to achieve a seamless translation table of lengths into angle values and vice versa (arcsin,...). Further information is on the sine page).

It should be noted that the geometric interpretation of the inner product is generally only used for [itex]\mathbb{R}^n[itex] with [itex]n \le 3[itex]. When working in higher dimensions, in spaces over other fields, or with modules, the only definition of the inner product is the following: [itex]\left \langle \mathbf{a}, \mathbf{b} \right \rangle = \sum_{i=1}^n a_ib_i[itex]

The dot product is particularly used in the calculation of net force. If b is a unit vector, then the dot product gives the projection of a in the direction b. In mechanics, this gives the component of a force in that direction.

Work is the dot product of force and displacement.

Properties

The definition has the following consequences. The dot product is commutative:

[itex] \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \;[itex].

Two non-zero vectors a and b are perpendicular if and only if a · b = 0. The dot product is bilinear:

[itex]
   \mathbf{a} \cdot (r\mathbf{b} +  \mathbf{c})


= r(\mathbf{a} \cdot \mathbf{b}) +(\mathbf{a} \cdot \mathbf{c}) \;[itex].

From these it follows directly that the dot product of two vectors a = [a1 a2 a3] and
b = [b1 b2 b3] given in coordinates can be computed particularly easily as

[itex] \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \; [itex],

or, using matrix multiplication and treating the vectors as 1-by-3 matrices,

[itex] \mathbf{a} \cdot \mathbf{b} = \mathbf{a b}^T \;[itex]

where bT denotes the transpose of the matrix b.

The dot product satisfies all the axioms of an inner product. In an abstract vector space, the notion of angle between the elements of the space can be defined in terms of the inner product.

Proof that the two forms of definition are equivalent

We have already shown that the theorem

[itex] \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \; [itex]

follows from the definition

[itex] \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \;[itex].

To prove that these are two equivalent ways of defining the dot product, we shall now instead use the former to derive the latter.

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

Consider a vector

[itex] \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k} \; [itex].

Repeated application of the Pythagorean theorem yields

[itex] v^2 = v_1^2 + v_2^2 + v_3^2 \;[itex].

But this is the same as

[itex] \mathbf{v} \cdot \mathbf{v} = v_1^2 + v_2^2 + v_3^2 \;[itex],

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

Lemma 1
[itex] \mathbf{v} \cdot \mathbf{v} = v^2 \; [itex].

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

[itex] \mathbf{c} \equiv \mathbf{a} - \mathbf{b} \;[itex],

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

[itex] c^2 = a^2 + b^2 - 2 ab \cos \theta \;[itex].

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

[itex]
 \mathbf{c} \cdot \mathbf{c}


= \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} - 2 ab \cos\theta \; [itex].                   (1) But as cab, we also have

[itex]
 \mathbf{c} \cdot \mathbf{c}


= (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \;[itex], which, according to the distributive law, expands to

[itex]
 \mathbf{c} \cdot \mathbf{c}


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

[itex]
 \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 \; [itex]. Subtracting a · a + b · b from both sides and dividing by −2 leaves

[itex] \mathbf{a} \cdot \mathbf{b} = ab \cos\theta \; [itex],

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