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Section 3.2 The 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 \(\mathbf u \cdot \mathbf v\text{.}\)

Definition 3.4. Dot Product.

\begin{equation} \mathbf u \cdot \mathbf v = u_1v_1+u_2v_2+u_3v_3.\tag{3.5} \end{equation}

Remark 3.5. 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.

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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Figure 3.8. The triangle with vertices \(A(1,-1)\text{,}\) \(B(1,1)\text{,}\) \(C(2,0)\text{,}\) together with the vectors \(\vec{CA}\) and \(\vec{CB}\) meeting at a right angle at \(C\text{.}\)

Example 3.9. The remaining angles of the triangle.

For the triangle of ExampleΒ 3.7 (shown again in FigureΒ 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*}

Subsection 3.2.1 Properties of the Dot Product

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