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Euclidean Spaces

The dot product is fundamentally a projection: the dot product of a vector with a unit vector is the projection of that vector in the direction given by the unit vector. This leads to the geometric formula
\[ {\bf u} \bullet {\bf v} = \vert {\bf u} \vert \cdot \vert {\bf v} \vert \cdot \cos\theta , \]
where θ is the angle between vectors u and v.

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 Augustin-Louis Cauchy    Viktor Yakovlevich Bunyakovsky

An immediate consequence of (1) is that the dot product of a vector with itself gives the square of the length, that is

\[ {\bf u} \bullet {\bf u} = \vert {\bf u} \vert^2 . \]
In particular, taking the “square” of any unit vector yields 1, for example
\[ {\bf i} \bullet {\bf i} = 1 \qquad \mox{and} \qquad {\bf j} \bullet {\bf j} = 1 , \tag{3} \]
where i and j are unit vectors along abscissa and ordinate. Furthermore, it follows immediately from the geometric definition that two vectors are orthogonal if and only if their dot product vanishes, that is,
\[ \mathbf{u} \perp \mathbf{v} \qquad \iff \qquad \mathbf{u} \bullet \mathbf{v} = 0 . \]
For instance,
\[ \mathbf{i} \perp \mathbf{j} \qquad \iff \qquad \mathbf{i} \bullet \mathbf{j} = 0 . \tag{5} \]
The geometry of an orthonormal basis is fully captured by these properties; each basis vector is normalized, which is (3), and each pair of vectors is orthogonal, which is (5). The components of a vector ~v in an orthonormal basis are just the dot products of v with each basis vector. For instance, in two dimensions
\[ \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} . \tag{5} \]
If u = uxi + uyj. then
\begin{align*} \mathbf{u} \bullet \mathbf{v} &= \left( u_x \mathbf{i} + u_y \mathbf{j} \right) \bullet \left( v_x \mathbf{i} + v_y \mathbf{j} \right) \\ &= u_x v_x + u_y v_y . \end{align*}
This computation clearly works for any orthonormal basis. A special case is the dot product of a vector with itself, which reduces to the Pythagorean theorem, for example,
\[ \mathbf{v} \bullet \mathbf{v} = \| \mathbf{v} \|^2 = v_x^2 + v_y^2 . \tag{8} \]
What happens if you don’t use an orthonormal basis? Consider Figure 2, in which u + v = w, or equivalently v = w − u. Then
\begin{align*} \mathbf{v} \bullet \mathbf{v} &= \left( \mathbf{w} - \mathbf{u} \right) \bullet \left( \mathbf{w} - \mathbf{u} \right) \\ &= \| \mathbf{w} \|^2 + \| \mathbf{u} \|^2 - 2\,\mathbf{w} \bullet \mathbf{u} . \end{align*}
This is equivalet to
\[ \| \mathbf{v} \|^2 = \| \mathbf{w} \|^2 + \| \mathbf{u} \|^2 - 2\,\| \mathbf{w} \| \,\| \mathbf{w} \| \,\cos\theta , \]
which is just the Law of Cosines! The Law of Cosines is usually used to derive the geometric form of the dot product (1) from the algebraic form (7), which is taken as the definition. Instead, by starting with geometry, the Law    
Example 1:    ■
End of Example 1

 

  1. What is the angle between the vectors i + j and i + 3j?
  2. What is the area of the quadrilateral with vertices at (1, 1), (4, 2), (3, 7) and (2, 3)?
  1. Vector addition
  2. Deay, T. and Manogue, C.A., he Geometry of the Dot and Cross Products, Journal of Online Mathematics and Its Applications 6.