Angle Vectors 676740

1. **Problem statement:** Find the angle between the vectors $$3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$$ and $$-4\mathbf{i} - \mathbf{j} + 3\mathbf{k}$$. 2. **Formula used:** The angle $$\theta$$ between two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ is given by $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$$ where $$\mathbf{a} \cdot \mathbf{b}$$ is the dot product and $$|\mathbf{a}|$$, $$|\mathbf{b}|$$ are the magnitudes of the vectors. 3. **Calculate the dot product:** $$\mathbf{a} = (3, 2, -2), \quad \mathbf{b} = (-4, -1, 3)$$ $$\mathbf{a} \cdot \mathbf{b} = 3 \times (-4) + 2 \times (-1) + (-2) \times 3 = -12 - 2 - 6 = -20$$ 4. **Calculate the magnitudes:** $$|\mathbf{a}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}$$ $$|\mathbf{b}| = \sqrt{(-4)^2 + (-1)^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}$$ 5. **Calculate $$\cos \theta$$:** $$\cos \theta = \frac{-20}{\sqrt{17} \times \sqrt{26}} = \frac{-20}{\sqrt{442}}$$ 6. **Calculate $$\theta$$:** $$\theta = \cos^{-1} \left( \frac{-20}{\sqrt{442}} \right)$$ Using a calculator, $$\theta \approx \cos^{-1}(-0.952) \approx 162.5^\circ$$ 7. **Interpretation:** The angle between the vectors is approximately $$162.5^\circ$$. **Final answer:** $$\boxed{162.5^\circ}$$