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To find the angle \(\theta\) between vectors \(\vec{a}\) and \(\vec{b}\), we can use the formula:

\[
\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}
\]

Where:
- \(\vec{a} = (2, -3, 1)\)
- \(\vec{b} = (-1, 0, 5)\)

### Step 1: Compute the dot product \(\vec{a} \cdot \vec{b}\)
The dot product of two vectors \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\) is calculated as:

\[
\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3
\]

Substitute the components of \(\vec{a}\) and \(\vec{b}\):

\[
\vec{a} \cdot \vec{b} = (2)(-1) + (-3)(0) + (1)(5) = -2 + 0 + 5 = 3
\]

### Step 2: Compute the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \)
The magnitude of a vector \(\vec{a} = (a_1, a_2, a_3)\) is calculated as:

\[
|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}
\]

For \(\vec{a} = (2, -3, 1)\):

\[
|\vec{a}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}
\]

For \(\vec{b} = (-1, 0, 5)\):

\[
|\vec{b}| = \sqrt{(-1)^2 + 0^2 + 5^2} = \sqrt{1 + 0 + 25} = \sqrt{26}
\]

### Step 3: Use the formula for \(\cos \theta\)
Substitute the values into the formula for \(\cos \theta\):

\[
\cos \theta = \frac{3}{\sqrt{14} \cdot \sqrt{26}} = \frac{3}{\sqrt{364}} = \frac{3}{19.08}
\]

\[
\cos \theta \approx 0.1572
\]

### Step 4: Solve for \(\theta\)
Now, take the inverse cosine (\(\cos^{-1}\)) of 0.1572 to find \(\theta\):

\[
\theta \approx \cos^{-1}(0.1572) \approx 81.0^\circ
\]

Thus, the angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) is approximately \(81.0^\circ\).

---

Would you like further clarification or details on any of these steps?

### Related Questions:
1. How do you calculate the dot product between two vectors geometrically?
2. What is the significance of the angle between two vectors in different dimensions?
3. Can two vectors have a zero angle between them, and what does that imply?
4. How does the magnitude of a vector affect its properties in physics?
5. How can this method be extended to find the angle between vectors in higher dimensions?

### Tip:
Remember, vectors with a dot product of zero are orthogonal, meaning they form a right angle.

If \(\vec{a} = (2, -3, 1)\) and \(\vec{b} = (-1, 0, 5)\), find \(\theta\), the angle between \(\vec{a}\) and \(\vec{b}\).

Learn how to calculate the angle between two 3D vectors, \(\vec{a} = (2, -3, 1)\) and \(\vec{b} = (-1, 0, 5)\), using vector dot product and magnitude formulas. Includes a detailed step-by-step solution suitable for Grades 10-12.

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