1. **Find the angle between vectors** $\mathbf{u} = \begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}$ and $\mathbf{v} = \begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}$ using the dot product formula. The dot product formula for angle $\theta$ is: $$\mathbf{u} \cdot \mathbf{v} = ||\mathbf{u}|| \, ||\mathbf{v}|| \cos \theta$$ Calculate $\mathbf{u} \cdot \mathbf{v}$: $$3 \cdot 1 + 2 \cdot 0 + 1 \cdot (-1) = 3 + 0 - 1 = 2$$ Calculate magnitudes: $$||\mathbf{u}|| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$$ $$||\mathbf{v}|| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$$ Use the formula to find $\cos \theta$: $$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \, ||\mathbf{v}||} = \frac{2}{\sqrt{14} \times \sqrt{2}} = \frac{2}{\sqrt{28}} = \frac{2}{2 \sqrt{7}} = \frac{1}{\sqrt{7}}$$ Find angle $\theta$: $$\theta = \cos^{-1} \left( \frac{1}{\sqrt{7}} \right)$$ 2. **Compute the projection of** $\mathbf{a} = \begin{bmatrix}2 \\ 3\end{bmatrix}$ **onto** $\mathbf{b} = \begin{bmatrix}1 \\ 4\end{bmatrix}$ and find the component of $\mathbf{a}$ in the direction of $\mathbf{b}$. Projection formula: $$\mathrm{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{b}||^2} \mathbf{b}$$ Calculate dot product $\mathbf{a} \cdot \mathbf{b}$: $$2 \times 1 + 3 \times 4 = 2 + 12 = 14$$ Calculate $||\mathbf{b}||^2$: $$1^2 + 4^2 = 1 + 16 = 17$$ Calculate projection: $$\mathrm{proj}_{\mathbf{b}} \mathbf{a} = \frac{14}{17} \begin{bmatrix}1 \\ 4\end{bmatrix} = \begin{bmatrix} \frac{14}{17} \\ \frac{56}{17} \end{bmatrix}$$ The component of $\mathbf{a}$ in the direction of $\mathbf{b}$ is the length of the projection vector: $$\left| \left| \mathrm{proj}_{\mathbf{b}} \mathbf{a} \right| \right| = \sqrt{\left( \frac{14}{17} \right)^2 + \left( \frac{56}{17} \right)^2} = \frac{1}{17} \sqrt{14^2 + 56^2} = \frac{1}{17} \sqrt{196 + 3136} = \frac{\sqrt{3332}}{17}$$ 3. **Determine whether vectors** $\mathbf{p} = \begin{bmatrix}1 \\ -2 \\ 1\end{bmatrix}$ **and** $\mathbf{q} = \begin{bmatrix}2 \\ 1 \\ -1\end{bmatrix}$ **are orthogonal**. Check if $\mathbf{p} \cdot \mathbf{q} = 0$: $$1 \times 2 + (-2) \times 1 + 1 \times (-1) = 2 - 2 - 1 = -1$$ Since the dot product is $-1 \neq 0$, the vectors are **not orthogonal**. **Final answers:** 1. Angle between $\mathbf{u}$ and $\mathbf{v}$ is: $$\theta = \cos^{-1} \left( \frac{1}{\sqrt{7}} \right)$$ 2. Projection of $\mathbf{a}$ onto $\mathbf{b}$: $$\mathrm{proj}_{\mathbf{b}} \mathbf{a} = \begin{bmatrix} \frac{14}{17} \\ \frac{56}{17} \end{bmatrix}$$ Component of $\mathbf{a}$ in direction of $\mathbf{b}$: $$\frac{\sqrt{3332}}{17}$$ 3. Vectors $\mathbf{p}$ and $\mathbf{q}$ are **not orthogonal**.