Dot Product - Multivariable Calculus

1. Consider vectors $\mathbf{p} = \langle 1, 2, -1 \rangle$ and $\mathbf{q} = \langle 3, -1, 2 \rangle$. If the angle between $\mathbf{p}$ and $\mathbf{r}$ is $60°$ and the angle between $\mathbf{q}$ and $\mathbf{r}$ is $120°$, where $||\mathbf{r}|| = 2$, what is $\mathbf{p} \cdot \mathbf{r} + \mathbf{q} \cdot \mathbf{r}$?
   Answer: $\sqrt{6} - \sqrt{14}$ Using the dot product formula $\mathbf{u} \cdot \mathbf{v} = ||\mathbf{u}|| ||\mathbf{v}|| \cos \theta$, we find: $\mathbf{p} \cdot \mathbf{r} = ||\mathbf{p}|| ||\mathbf{r}|| \cos 60° = \sqrt{6} \cdot 2 \cdot \frac{1}{2} = \sqrt{6}$ and $\mathbf{q} \cdot \mathbf{r} = ||\mathbf{q}|| ||\mathbf{r}|| \cos 120° = \sqrt{14} \cdot 2 \cdot (-\frac{1}{2}) = -\sqrt{14}$. Therefore, $\mathbf{p} \cdot \mathbf{r} + \mathbf{q} \cdot \mathbf{r} = \sqrt{6} - \sqrt{14}$. Here $||\mathbf{p}|| = \sqrt{1 + 4 + 1} = \sqrt{6}$ and $||\mathbf{q}|| = \sqrt{9 + 1 + 4} = \sqrt{14}$. Option B incorrectly doubles both terms, C reverses the signs, and D both reverses signs and doubles.
2. Two vectors $\mathbf{u}$ and $\mathbf{v}$ satisfy $||\mathbf{u}|| = 3$, $||\mathbf{v}|| = 4$, and $||\mathbf{u} + \mathbf{v}|| = 6$. What is the length of the projection of $\mathbf{u} - \mathbf{v}$ onto $\mathbf{u} + \mathbf{v}$?
   Answer: $\frac{7}{6}$ First, we find $\mathbf{u} \cdot \mathbf{v}$ using $||\mathbf{u} + \mathbf{v}||^2 = ||\mathbf{u}||^2 + ||\mathbf{v}||^2 + 2\mathbf{u} \cdot \mathbf{v}$. We have $36 = 9 + 16 + 2\mathbf{u} \cdot \mathbf{v}$, so $\mathbf{u} \cdot \mathbf{v} = \frac{11}{2}$. The projection length is $\frac{|(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})|}{||\mathbf{u} + \mathbf{v}||}$. We compute $(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = ||\mathbf{u}||^2 - ||\mathbf{v}||^2 = 9 - 16 = -7$. Therefore, the projection length is $\frac{|-7|}{6} = \frac{7}{6}$. Option A uses the wrong dot product calculation, C and D incorrectly compute the denominator.

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