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Tentukanlah sudut diantara vector-vektor berikut, jika berlaku hasil kali titik. selesaikan menggunakan cara ruang hasil kali dalam

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!

Learn how to find the angle between vectors using the dot product. This solution demonstrates step-by-step calculations for vectors (-1, 5, 2) and (2, 4, -9), and (2, 1, -1) and (4, 0, 0). Suitable for high school students studying vector mathematics.

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