Test Objectives
• Demonstrate the ability to work with a linear combination of unit vectors
• Demonstrate the ability to find the dot product of two vectors
• Demonstrate the ability to find the angle between two vectors
• Demonstrate the ability to determine if two vectors are orthogonal
Dot Product & Angle Between Vectors Practice Test:

#1:

Instructions: Find the unit vector u in the direction of v.

$$a)\hspace{.1em}\overrightarrow{v}=-4\hat{\text{i}}+ 8\hat{\text{j}}$$

$$b)\hspace{.1em}\overrightarrow{v}=\langle 30, 40 \rangle$$

#2:

Instructions: Find the dot product of the vectors u and v.

$$a)\hspace{.1em}$$ $$\overrightarrow{u}=\langle -8, -3 \rangle$$ $$\overrightarrow{v}=\langle 6, 6 \rangle$$

$$b)\hspace{.1em}$$ $$\overrightarrow{u}=-9\hat{\text{i}}- 2\hat{\text{j}}$$ $$\overrightarrow{v}=-3\hat{\text{i}}- 8\hat{\text{j}}$$

#3:

Instructions: Find the measure of the angle θ between vectors u and v geometrically.

$$a)\hspace{.1em}$$ $$\overrightarrow{u}=\langle 8, 3 \rangle$$ $$\overrightarrow{v}=\langle 2, 7 \rangle$$

Hint: Sketch the vectors u and v in standard position. Form a triangle using u - v. Use the law of cosines to find the angle between u and v.

Instructions: Find the measure of the angle θ between vectors u and v.

$$b)\hspace{.1em}$$ $$\overrightarrow{u}=\langle 2, -4 \rangle$$ $$\overrightarrow{v}=\langle 4, -5 \rangle$$

$$c)\hspace{.1em}$$ $$\overrightarrow{u}=\langle 2, 9 \rangle$$ $$\overrightarrow{v}=\langle 8, -3 \rangle$$

#4:

Instructions: Find the measure of the angle θ between vectors u and v.

$$a)\hspace{.1em}$$ $$\overrightarrow{u}=-\hat{\text{i}}- 3\hat{\text{j}}$$ $$\overrightarrow{v}=3\hat{\text{i}}+ 8\hat{\text{j}}$$

$$b)\hspace{.1em}$$ $$\overrightarrow{u}=-3\hat{\text{i}}+ 5\hat{\text{j}}$$ $$\overrightarrow{v}=3\hat{\text{i}}- 8\hat{\text{j}}$$

#5:

Instructions: Determine if the vectors u and v are orthogonal.

$$a)\hspace{.1em}$$ $$\overrightarrow{u}=\langle -5, -2 \rangle$$ $$\overrightarrow{v}=\langle -2, -5 \rangle$$

$$b)\hspace{.1em}$$ $$\overrightarrow{u}=9\hat{\text{i}}- 4\hat{\text{j}}$$ $$\overrightarrow{v}=-8\hat{\text{i}}- 18\hat{\text{j}}$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.1em}\hat{\text{u}}=-\frac{\sqrt{5}}{5}\hat{\text{i}}+ \frac{2 \sqrt{5}}{5}\hat{\text{j}}$$

$$b)\hspace{.1em}\hat{\text{u}}=\left\langle \frac{3}{5}, \frac{4}{5}\right\rangle$$

#2:

Solutions:

$$a)\hspace{.1em}\overrightarrow{u}\cdot \overrightarrow{v}=-66$$

$$b)\hspace{.1em}\overrightarrow{u}\cdot \overrightarrow{v}=43$$

#3:

Solutions:

$$a)$$ $$\left| \hspace{.2em}\overrightarrow{u}- \overrightarrow{v}\hspace{.2em}\right|=2\sqrt{13}$$ $$\left| \hspace{.2em}\overrightarrow{u}\hspace{.2em}\right|=\sqrt{73}$$ $$\left| \hspace{.2em}\overrightarrow{v}\hspace{.2em}\right|=\sqrt{53}$$ Law of Cosines: $$\left| \hspace{.2em}\overrightarrow{u}- \overrightarrow{v}\hspace{.2em}\right|^2=\left| \hspace{.2em}\overrightarrow{u}\hspace{.2em}\right|^2 + \left| \hspace{.2em}\overrightarrow{v}\hspace{.2em}\right|^2 - 2\left| \hspace{.2em}\overrightarrow{u}\hspace{.2em}\right| \left| \hspace{.2em}\overrightarrow{v}\hspace{.2em}\right| \cdot \text{cos}\hspace{.2em}θ \hspace{.2em}$$ $$(2\sqrt{13})^2=(\sqrt{73})^2 + (\sqrt{53})^2 - 2(\sqrt{73})(\sqrt{53}) \cdot \text{cos}\hspace{.2em}θ \hspace{.2em}$$ $$52=73 + 53 - 2(\sqrt{73})(\sqrt{53}) \cdot \text{cos}\hspace{.2em}θ \hspace{.2em}$$ Solve for cos θ: $$\text{cos}\hspace{.2em}θ=\frac{-74}{- 2(\sqrt{73})(\sqrt{53})}$$ $$\text{cos}\hspace{.2em}θ=\frac{37}{\sqrt{73}\cdot \sqrt{53}}$$ $$θ ≈ 53.5°$$

$$b)\hspace{.1em}θ ≈ 12.09°$$

$$c)\hspace{.1em}θ ≈ 98.03°$$

#4:

Solutions:

$$a)\hspace{.1em}θ ≈ 177.88°$$

$$b)\hspace{.1em}θ ≈ 169.59°$$

#5:

Solutions:

$$a)\hspace{.1em}\text{Not Orthogonal}$$

$$b)\hspace{.1em}\text{Orthogonal}$$