Cosine Formula for Dot Product

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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 1

There are two cases:

$(1): \quad \mathbf v$ and $\mathbf w$ are not scalar multiples of each other
$(2): \quad \mathbf v$ and $\mathbf w$ are scalar multiples of each other.


Case 1

AngleBetweenTwoVectors.png

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 $\norm {\mathbf u}$
\(\ds \) \(=\) \(\ds \norm {\mathbf v} \norm {\mathbf w} \cos \theta\) $\sgn \paren c = \cos \theta$

$\blacksquare$


Proof 2

Let $\mathbf v$ and $\mathbf w$ be considered to be embedded in a Cartesian plane $\CC$.

By Dot Product is Invariant under Coordinate Rotation, we may rotate $\CC$ arbitrarily, and $\mathbf v \cdot \mathbf w$ will not change.

So, let us rotate $\CC$ to $\CC'$ such that the $x$-axis is parallel to $\mathbf v$.

Hence $\mathbf v$ can be expressed as:

\(\ds \mathbf v\) \(=\) \(\ds \norm {\mathbf v} \mathbf i + 0 \mathbf j + 0 \mathbf k\)
\(\ds \mathbf w\) \(=\) \(\ds \norm {\mathbf w} \cos \theta \mathbf i + \norm {\mathbf w} \sin \theta \mathbf j + 0 \mathbf k\)

Hence: by definition of dot product:

\(\ds \mathbf v \cdot \mathbf w\) \(=\) \(\ds \norm {\mathbf v} \times \norm {\mathbf w} \cos \theta + 0 \times \norm {\mathbf w} \sin \theta + 0 \times 0\) Definition of Dot Product
\(\ds \) \(=\) \(\ds \norm {\mathbf v} \norm {\mathbf w} \cos \theta\)

Hence the result.

$\blacksquare$


Also see