1. **State the problem:** Given vectors $\mathbf{A} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$ and $\mathbf{B} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$, find the angle $\theta$ between $\mathbf{A}$ and $\mathbf{B}$. 2. **Calculate the dot product $\mathbf{A} \cdot \mathbf{B}$:** $$\mathbf{A} \cdot \mathbf{B} = 2 \times 1 + 3 \times (-2) + 1 \times 2 = 2 - 6 + 2 = -2$$ 3. **Calculate the magnitudes $|\mathbf{A}|$ and $|\mathbf{B}|$:** $$|\mathbf{A}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$$ $$|\mathbf{B}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$ 4. **Calculate $\cos \theta$ using the dot product formula:** $$\cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{-2}{3 \sqrt{14}}$$ 5. **Calculate the angle $\theta$:** $$\theta = \cos^{-1} \left( \frac{-2}{3 \sqrt{14}} \right)$$ Using a calculator, compute the value inside the inverse cosine: $$\frac{-2}{3 \sqrt{14}} \approx \frac{-2}{3 \times 3.7417} = \frac{-2}{11.225} \approx -0.1782$$ Then, $$\theta = \cos^{-1}(-0.1782) \approx 100.3^\circ$$ 6. **Why your calculator shows 1.7499:** Your calculator likely returned the angle in radians, not degrees. To convert radians to degrees: $$1.7499 \text{ radians} \times \frac{180}{\pi} \approx 100.3^\circ$$ 7. **Check if vectors are perpendicular:** Vectors are perpendicular if $\mathbf{A} \cdot \mathbf{B} = 0$. Here, $\mathbf{A} \cdot \mathbf{B} = -2 \neq 0$, so they are not perpendicular. 8. **Find the projection of $\mathbf{A}$ on $\mathbf{B}$:** $$\text{proj}_{\mathbf{B}} \mathbf{A} = \left( \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2} \right) \mathbf{B} = \left( \frac{-2}{9} \right)(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) = -\frac{2}{9} \mathbf{i} + \frac{4}{9} \mathbf{j} - \frac{4}{9} \mathbf{k}$$ **Final answer:** - Angle between $\mathbf{A}$ and $\mathbf{B}$ is approximately $100.3^\circ$. - Vectors are not perpendicular. - Projection of $\mathbf{A}$ on $\mathbf{B}$ is $-\frac{2}{9} \mathbf{i} + \frac{4}{9} \mathbf{j} - \frac{4}{9} \mathbf{k}$.