Watch the video below for a proof of the law of cosines using vectors.
Section 3.1 The Dot Product and the Angle between Two Vectors
Consider two vectors \(\mathbf u\) and \(\mathbf v\text{,}\) and let \(\mathbf w = \mathbf u - \mathbf v\text{,}\) as shown in Figure 3.1. By the law of cosines,
\begin{equation}
\|\mathbf w\|^2 = \|\mathbf u\|^2 + \|\mathbf v\|^2 - 2\|\mathbf u\|\|\mathbf v\|\cos\theta.\tag{3.1}
\end{equation}
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Proof.
Writing \(\mathbf u = \langle u_1, u_2, u_3\rangle\) and \(\mathbf v = \langle v_1, v_2, v_3\rangle\text{,}\) note that \(\mathbf w = \mathbf u - \mathbf v\text{,}\) and hence
\begin{equation}
\begin{aligned}
\|\mathbf w\|^2 \amp = \|\mathbf u - \mathbf v\|^2 \\
\amp = (u_1-v_1)^2 + (u_2-v_2)^2 + (u_3-v_3)^2 \\
\amp = (u_1^2+u_2^2+u_3^2) + (v_1^2+v_2^2+v_3^2) \\
\amp \qquad - 2(u_1v_1+u_2v_2+u_3v_3).
\end{aligned}\tag{3.2}
\end{equation}
Also, from the law of cosines in (3.1) we have
\begin{equation}
\|\mathbf w\|^2 = (u_1^2+u_2^2+u_3^2) + (v_1^2+v_2^2+v_3^2) - 2\|\mathbf u\|\|\mathbf v\|\cos\theta.\tag{3.3}
\end{equation}
From (3.2) and (3.3), it follows that \(u_1v_1+u_2v_2+u_3v_3 = \|\mathbf u\|\|\mathbf v\|\cos\theta\text{,}\) and therefore the angle between the two vectors \(\mathbf u\) and \(\mathbf v\) can be evaluated using the following equation.
Theorem 3.3 Angle between Two Vectors.
Let \(\mathbf u = \langle u_1, u_2, u_3\rangle\) and \(\mathbf v = \langle v_1, v_2, v_3\rangle\) be two nonzero vectors, and let \(\theta\text{,}\) with \(0 \le \theta \le \pi\text{,}\) be the angle between them. Then
\begin{equation}
\begin{aligned}
\cos\theta \amp = \frac{u_1v_1+u_2v_2+u_3v_3}{\|\mathbf u\|\|\mathbf v\|} \\
\iff\quad \theta \amp = \cos^{-1}\left(\frac{u_1v_1+u_2v_2+u_3v_3}{\|\mathbf u\|\|\mathbf v\|}\right).
\end{aligned}\tag{3.4}
\end{equation}
Definition 3.4 Dot Product.
The term \(u_1v_1+u_2v_2+u_3v_3\) in the numerator of (3.4) is known as the dot product of the two vectors \(\mathbf u = \langle u_1,u_2,u_3\rangle\) and \(\mathbf v = \langle v_1,v_2,v_3\rangle\text{,}\) and is denoted \(\mathbf u \cdot \mathbf v\text{.}\)
\begin{equation}
\mathbf u \cdot \mathbf v = u_1v_1+u_2v_2+u_3v_3.\tag{3.5}
\end{equation}
Example 3.5 Computing a dot product.
Let \(\mathbf u = \langle 1,2,3\rangle\) and \(\mathbf v = \langle -2,0,4\rangle\text{.}\) Compute \(\mathbf u \cdot \mathbf v\text{,}\) and use it to find the angle between \(\mathbf u\) and \(\mathbf v\text{.}\)
Solution.
Multiply the corresponding components and add:
\begin{equation*}
\mathbf u \cdot \mathbf v = (1)(-2) + (2)(0) + (3)(4) = -2 + 0 + 12 = 10.
\end{equation*}
Since the dot product is positive, the angle between the two vectors is acute. To find it, use (3.4):
\begin{equation*}
\begin{aligned}
\|\mathbf u\| \amp = \sqrt{1+4+9} = \sqrt{14} \\
\|\mathbf v\| \amp = \sqrt{4+0+16} = \sqrt{20} \\
\cos\theta \amp = \frac{\mathbf u \cdot \mathbf v}{\|\mathbf u\|\|\mathbf v\|}
= \frac{10}{\sqrt{14}\,\sqrt{20}} \approx 0.5976 \\
\theta \amp = \cos^{-1}(0.5976) \approx 0.930 \text{ radians} \approx 53.3^{\circ}.
\end{aligned}
\end{equation*}
Remark 3.6 Orthogonal vectors.
If the two vectors \(\mathbf u\) and \(\mathbf v\) are orthogonal, then \(\theta = \pi/2\text{,}\) which means \(\mathbf u \cdot \mathbf v = 0\text{.}\) Conversely, if \(\mathbf u\) and \(\mathbf v\) are two vectors such that \(\mathbf u \cdot \mathbf v = 0\text{,}\) then \(\mathbf u\) and \(\mathbf v\) are orthogonal.
\begin{equation}
\mathbf u \cdot \mathbf v = 0 \iff \mathbf u \perp \mathbf v.\tag{3.6}
\end{equation}
Example 3.7 A right triangle detected with the dot product.
Consider the triangle with vertices \(A(1,-1)\text{,}\) \(B(1,1)\text{,}\) and \(C(2,0)\text{.}\) Show that this is a right triangle using the dot product.
Solution.
\begin{equation*}
\begin{aligned}
\vec{CA} \amp = \langle 1-2,\,-1-0\rangle = \langle -1,-1\rangle \\
\vec{CB} \amp = \langle 1-2,\,1-0\rangle = \langle -1,1\rangle \\
\vec{CA}\cdot\vec{CB} \amp = 1 - 1 = 0 \implies \vec{CA} \perp \vec{CB}.
\end{aligned}
\end{equation*}
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Example 3.9 The remaining angles of the triangle.
For the triangle of Example 3.7 (shown again in Figure 3.8), calculate the remaining angles.
Solution.
For the angle between the sides \(\vec{AC}\) and \(\vec{AB}\text{,}\) we have
\begin{equation*}
\begin{aligned}
\vec{AC} \amp = \langle 2-1,\,0+1\rangle = \langle 1,1\rangle \implies \|\vec{AC}\| = \sqrt2 \\
\vec{AB} \amp = \langle 1-1,\,1+1\rangle = \langle 0,2\rangle \implies \|\vec{AB}\| = 2 \\
\vec{AB}\cdot\vec{AC} \amp = 0 + 2 = 2 \\
\cos\theta \amp = \frac{\vec{AB}\cdot\vec{AC}}{\|\vec{AB}\|\|\vec{AC}\|}
= \frac{2}{2\sqrt2} = \frac{1}{\sqrt2} \implies \theta = \frac{\pi}{4}.
\end{aligned}
\end{equation*}
Properties of the Dot Product.
Fact 3.10.
If \(\mathbf u\text{,}\) \(\mathbf v\text{,}\) and \(\mathbf w\) are any vectors and \(c\) is a scalar, then
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\(\displaystyle \mathbf u \cdot \mathbf v = \mathbf v \cdot \mathbf u\)
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\(\displaystyle (c\mathbf u)\cdot \mathbf v = \mathbf u \cdot (c\mathbf v) = c(\mathbf u \cdot \mathbf v)\)
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\(\displaystyle \mathbf u \cdot (\mathbf v + \mathbf w) = \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf w\)
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\(\displaystyle \mathbf u \cdot \mathbf u = \|\mathbf u\|^2\)
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\(\displaystyle \mathbf 0 \cdot \mathbf u = 0\)
Proof.
Write \(\mathbf u = \langle u_1,u_2,u_3\rangle\text{,}\) \(\mathbf v = \langle v_1,v_2,v_3\rangle\text{,}\) and \(\mathbf w = \langle w_1,w_2,w_3\rangle\text{.}\) Each property follows from (3.5) together with the corresponding property of real numbers.
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Multiplication of real numbers is commutative, so\begin{equation*} \mathbf u \cdot \mathbf v = u_1v_1+u_2v_2+u_3v_3 = v_1u_1+v_2u_2+v_3u_3 = \mathbf v \cdot \mathbf u. \end{equation*}
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Since \(c\mathbf u = \langle cu_1,cu_2,cu_3\rangle\text{,}\)\begin{equation*} \begin{aligned} (c\mathbf u)\cdot \mathbf v \amp = cu_1v_1+cu_2v_2+cu_3v_3 \\ \amp = c(u_1v_1+u_2v_2+u_3v_3) = c(\mathbf u \cdot \mathbf v), \end{aligned} \end{equation*}and the same computation applied to \(c\mathbf v = \langle cv_1,cv_2,cv_3\rangle\) gives \(\mathbf u \cdot (c\mathbf v) = c(\mathbf u \cdot \mathbf v)\text{.}\)
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Adding componentwise and then distributing,\begin{equation*} \begin{aligned} \mathbf u \cdot (\mathbf v + \mathbf w) \amp = u_1(v_1+w_1) + u_2(v_2+w_2) + u_3(v_3+w_3) \\ \amp = (u_1v_1+u_2v_2+u_3v_3) + (u_1w_1+u_2w_2+u_3w_3) \\ \amp = \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf w. \end{aligned} \end{equation*}
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Taking \(\mathbf v = \mathbf u\) in (3.5), and recalling that \(\|\mathbf u\| = \sqrt{u_1^2+u_2^2+u_3^2}\text{,}\)\begin{equation*} \mathbf u \cdot \mathbf u = u_1^2+u_2^2+u_3^2 = \left(\sqrt{u_1^2+u_2^2+u_3^2}\right)^2 = \|\mathbf u\|^2. \end{equation*}
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Every component of \(\mathbf 0\) is zero, so \(\mathbf 0 \cdot \mathbf u = 0u_1+0u_2+0u_3 = 0\text{.}\)
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