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# Chapter 7. Vectors ## 7.4 The Dot Product ### Angle Between Two Vectors **Definition:** If $\vec{a}$ and $\vec{b}$ are two nonzero vectors, the angle $\theta$ between them is given by $\qquad \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$, $\qquad$ $0 \le \theta \le \pi$ **E...

# Chapter 7. Vectors ## 7.4 The Dot Product ### Angle Between Two Vectors **Definition:** If $\vec{a}$ and $\vec{b}$ are two nonzero vectors, the angle $\theta$ between them is given by $\qquad \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$, $\qquad$ $0 \le \theta \le \pi$ **Example 1:** Find the angle between the vectors $\vec{a} = \langle 4, 3 \rangle$ and $\vec{b} = \langle 2, -1 \rangle$. **Solution:** $\qquad \vec{a} \cdot \vec{b} = (4)(2) + (3)(-1) = 8 - 3 = 5$ $\qquad |\vec{a}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ $\qquad |\vec{b}| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$ So, $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{5}{5 \sqrt{5}} = \frac{1}{\sqrt{5}}$ Thus, $\theta = \cos^{-1} (\frac{1}{\sqrt{5}}) \approx 63.4^\circ$. ### Direction Angles and Direction Cosines Suppose that $\vec{a}$ is a three-dimensional vector and let $\alpha, \beta, \gamma$ be the angles that $\vec{a}$ makes with the positive $x, y,$ and $z$ axes. These angles are the **direction angles** of $\vec{a}$ and their cosines are the **direction cosines** of $\vec{a}$. Using the formula to determine the angle between two vectors, we have $\qquad \cos \alpha = \frac{\vec{a} \cdot \vec{i}}{|\vec{a}| |\vec{i}|} = \frac{a_1}{|\vec{a}|}$, $\qquad \cos \beta = \frac{\vec{a} \cdot \vec{j}}{|\vec{a}| |\vec{j}|} = \frac{a_2}{|\vec{a}|}$, $\qquad \cos \gamma = \frac{\vec{a} \cdot \vec{k}}{|\vec{a}| |\vec{k}|} = \frac{a_3}{|\vec{a}|}$ By squaring the direction cosines and adding, we have $\qquad \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \frac{a_1^2 + a_2^2 + a_3^2}{|\vec{a}|^2} = \frac{|\vec{a}|^2}{|\vec{a}|^2} = 1$ We can also write $\qquad \vec{a} = \langle a_1, a_2, a_3 \rangle = \langle |\vec{a}| \cos \alpha, |\vec{a}| \cos \beta, |\vec{a}| \cos \gamma \rangle = |\vec{a}| \langle \cos \alpha, \cos \beta, \cos \gamma \rangle$ Therefore $\qquad \frac{1}{|\vec{a}|} \vec{a} = \langle \cos \alpha, \cos \beta, \cos \gamma \rangle$ which says that the direction cosines of $\vec{a}$ are the components of the unit vector in the direction of $\vec{a}$. ### Projections **Scalar Projection:** The scalar projection of $\vec{b}$ onto $\vec{a}$ (also called the component of $\vec{b}$ along $\vec{a}$) is $\qquad \text{comp}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$ **Vector Projection:** The vector projection of $\vec{b}$ onto $\vec{a}$ is $\qquad \text{proj}_{\vec{a}} \vec{b} = (\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}) \frac{\vec{a}}{|\vec{a}|} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \vec{a}$ **Example 2:** Find the scalar and vector projections of $\vec{b} = \langle 4, 8 \rangle$ onto $\vec{a} = \langle 1, 2 \rangle$. **Solution:** $\qquad \text{comp}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \frac{(1)(4) + (2)(8)}{\sqrt{1^2 + 2^2}} = \frac{4 + 16}{\sqrt{5}} = \frac{20}{\sqrt{5}} = 4 \sqrt{5}$ $\qquad \text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \vec{a} = \frac{20}{5} \langle 1, 2 \rangle = 4 \langle 1, 2 \rangle = \langle 4, 8 \rangle$