To solve the problem of finding the angle between the vectors using the dot product (inner product in Euclidean space), we use the following formula:

$\cos \theta = \frac{\vec{P} \cdot \vec{Q}}{\|\vec{P}\| \|\vec{Q}\|}$

Where:

• $\vec{P} \cdot \vec{Q}$ is the dot product of vectors $P$ and $Q$.
• $\|\vec{P}\|$ and $\|\vec{Q}\|$ are the magnitudes of $P$ and $Q$.
• $\theta$ is the angle between the vectors.

Let's solve for both cases.

Case (a): $P = (-1, 5, 2)$ and $Q = (2, 4, -9)$

1. Dot Product: $\vec{P} \cdot \vec{Q} = (-1)(2) + (5)(4) + (2)(-9)$ Calculate: $\vec{P} \cdot \vec{Q} = -2 + 20 - 18 = 0$

2. Magnitudes:

   • $\|\vec{P}\| = \sqrt{(-1)^2 + 5^2 + 2^2} = \sqrt{1 + 25 + 4} = \sqrt{30}$
   • $\|\vec{Q}\| = \sqrt{(2)^2 + 4^2 + (-9)^2} = \sqrt{4 + 16 + 81} = \sqrt{101}$

3. Angle: $\cos \theta = \frac{\vec{P} \cdot \vec{Q}}{\|\vec{P}\| \|\vec{Q}\|} = \frac{0}{\sqrt{30} \cdot \sqrt{101}} = 0$ Thus, $\theta = \arccos(0) = 90^\circ$.

Case (b): $P = (2, 1, -1)$ and $Q = (4, 0, 0)$

1. Dot Product: $\vec{P} \cdot \vec{Q} = (2)(4) + (1)(0) + (-1)(0)$ Calculate: $\vec{P} \cdot \vec{Q} = 8 + 0 + 0 = 8$

2. Magnitudes:

   • $\|\vec{P}\| = \sqrt{(2)^2 + (1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$
   • $\|\vec{Q}\| = \sqrt{(4)^2 + (0)^2 + (0)^2} = \sqrt{16} = 4$

3. Angle: $\cos \theta = \frac{\vec{P} \cdot \vec{Q}}{\|\vec{P}\| \|\vec{Q}\|} = \frac{8}{\sqrt{6} \cdot 4} = \frac{8}{4\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}$ $\theta = \arccos\left(\frac{\sqrt{6}}{3}\right)$

Final Results:

• (a): $\theta = 90^\circ$
• (b): $\theta = \arccos\left(\frac{\sqrt{6}}{3}\right)$ or approximately $35.26^\circ$.

Let me know if you'd like more details or clarifications!

Relative Questions:

1. How is the dot product of two vectors calculated geometrically?
2. What is the significance of $\cos \theta = 0$ in the context of vector angles?
3. How do the magnitudes of vectors affect their inner product?
4. Why is the $\arccos$ function used to calculate the angle between vectors?
5. Can vectors in higher dimensions use the same formula for angles?

Tip: Always ensure that the dot product formula is applied correctly by confirming the dimensions of the vectors!