Cosine Formula for Dot Product
Theorem
Let $\mathbf v,\mathbf w$ be two non-zero vectors in $\R^n$.
The dot product of $\mathbf v$ and $\mathbf w$ can be calculated by:
- $\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
where:
- $\norm {\, \cdot \,}$ denotes vector length and
- $\theta$ is the angle between $\mathbf v$ and $\mathbf w$.
Proof
There are two cases, the first where the two vectors are not scalar multiples of each other, and the second where they are.
Case 1
Let the two vectors $\mathbf v$ and $\mathbf w$ not be scalar multiples of each other.
Then by the definition of angle between vectors, we have $\theta$ defined as in the triangle as shown above.
(Note that from Angle Between Non-Zero Vectors Always Defined, such a triangle is guaranteed to exist).
By the Law of Cosines:
- $\norm {\mathbf v - \mathbf w}^2 = \norm {\mathbf w}^2 + \norm {\mathbf v}^2 - 2 \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
Now, observe that:
| \(\ds \norm {\mathbf v - \mathbf w}^2\) | \(=\) | \(\ds \paren {\mathbf v - \mathbf w} \cdot \paren {\mathbf v - \mathbf w}\) | Dot Product of Vector with Itself | |||||||||||
| \(\ds \) | \(=\) | \(\ds \mathbf v \cdot \mathbf v - 2 \paren {\mathbf v \cdot \mathbf w} + \mathbf w \cdot \mathbf w\) | Properties of Dot Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\mathbf v}^2 + \norm {\mathbf w}^2 - 2 \paren {\mathbf v \cdot \mathbf w}\) | Dot Product of Vector with Itself |
Equating these two expressions for $\norm {\mathbf v - \mathbf w}^2$ gives:
| \(\ds \norm {\mathbf w}^2 + \norm {\mathbf v}^2 - 2 \norm {\mathbf v} \norm {\mathbf w} \cos \theta\) | \(=\) | \(\ds \norm {\mathbf v}^2 + \norm {\mathbf w}^2 - 2 \paren {\mathbf v \cdot \mathbf w}\) | ||||||||||||
| \(\ds -2 \paren {\norm {\mathbf v} \norm {\mathbf w} \cos \theta}\) | \(=\) | \(\ds -2 \paren {\mathbf v \cdot \mathbf w}\) | ||||||||||||
| \(\ds \norm {\mathbf v} \norm {\mathbf w} \cos \theta\) | \(=\) | \(\ds \mathbf v \cdot \mathbf w\) |
which is exactly the desired result.
$\Box$
Case 2
Let $\mathbf v = \paren {v_1, v_2, \ldots, v_n}$ and $\mathbf w = \paren {w_1, w_2, \ldots, w_n}$.
Without loss of generality, let $\mathbf v = c \mathbf w$, where $c$ is some scalar.
If $c > 0$, then by the definition of angle between vectors:
- $\theta = 0 \implies \cos \theta = 1$
If $c < 0$, then by the definition of angle between vectors:
- $\theta = \pi \implies \cos \theta = -1$
(Note that $c$ cannot be $0$ because we have stipulated $\mathbf v$ and $\mathbf w$ to be non-zero).
Then:
| \(\ds \mathbf v \cdot \mathbf w\) | \(=\) | \(\ds \paren {c \mathbf w} \cdot \mathbf w\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds c \paren {\mathbf w \cdot \mathbf w}\) | Properties of Dot Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds c \paren {\norm {\mathbf w}^2}\) | Dot Product of Vector with Itself | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {c \norm {\mathbf w} } \norm {\mathbf w}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds c \sqrt {\sum_{i \mathop = 1}^n w_i^2} \norm {\mathbf w}\) | Definition of Vector Length in $\R^n$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sgn \paren c \sqrt{c^2 \sum_{i \mathop = 1}^n w_i^2} \norm {\mathbf w}\) | where $\sgn \paren c$ is the signum function, that is the algebraic sign of $c$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sgn \paren c \sqrt {\sum_{i \mathop = 1}^n \paren {c w_i}^2} \norm {\mathbf w}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sgn \paren c \sqrt {\sum_{i \mathop = 1}^n v_i^2} \norm {\mathbf w}\) | by hypothesis | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sgn \paren c \norm {\mathbf v} \norm {\mathbf w}\) | Definition of $\norm {\mathbf u}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\mathbf v} \norm {\mathbf w} \cos \theta\) | $\sgn \paren c = \cos \theta$ |
$\blacksquare$
Also see
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.