To find the angle $\theta$ between two vectors $\mathbf{a} = (2, -1, 1)$ and $\mathbf{b} = (-1, 3, 5)$, we use the formula:

$\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$

Step 1: Compute the dot product $\mathbf{a} \cdot \mathbf{b}$

$\mathbf{a} \cdot \mathbf{b} = (2)(-1) + (-1)(3) + (1)(5) = -2 - 3 + 5 = 0$

Step 2: Compute the magnitudes $\|\mathbf{a}\|$ and $\|\mathbf{b}\|$

$\|\mathbf{a}\| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ $\|\mathbf{b}\| = \sqrt{(-1)^2 + (3)^2 + (5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$

Step 3: Substitute into the formula

$\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \frac{0}{\sqrt{6} \cdot \sqrt{35}} = 0$

Step 4: Solve for $\theta$

Since $\cos\theta = 0$, the angle $\theta$ is: $\theta = \cos^{-1}(0) = 90^\circ$

Final Answer:

The angle between the vectors is $90^\circ$, meaning the vectors are perpendicular.

Would you like details on any part of this process?

Follow-up Questions:

1. What is the significance of the dot product being zero?
2. How can this method be extended to higher dimensions?
3. Why does $\cos\theta = 0$ imply perpendicularity?
4. How are vector magnitudes used in real-world applications?
5. Can the angle between vectors be greater than $90^\circ$?

Tip:

Always check the dot product first; if it's zero, the vectors are perpendicular, saving extra computation.