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To find the angle between the vectors \( \mathbf{u} = \mathbf{i} + \sqrt{7} \mathbf{j} \) and \( \mathbf{v} = -\mathbf{i} + 4\mathbf{j} \), we use the formula for the angle \( \theta \) between two vectors:

\[
\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}
\]

### Step 1: Compute the dot product \( \mathbf{u} \cdot \mathbf{v} \)
The dot product of two vectors \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \) is given by:

\[
\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2
\]

Here:
- \( \mathbf{u} = (1, \sqrt{7}) \)
- \( \mathbf{v} = (-1, 4) \)

Substitute into the formula:

\[
\mathbf{u} \cdot \mathbf{v} = (1)(-1) + (\sqrt{7})(4)
\]
\[
\mathbf{u} \cdot \mathbf{v} = -1 + 4\sqrt{7}
\]

### Step 2: Find the magnitudes \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \)
The magnitude of a vector \( \mathbf{u} = (u_1, u_2) \) is:

\[
\|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2}
\]

For \( \mathbf{u} = (1, \sqrt{7}) \):

\[
\|\mathbf{u}\| = \sqrt{1^2 + (\sqrt{7})^2} = \sqrt{1 + 7} = \sqrt{8}
\]

For \( \mathbf{v} = (-1, 4) \):

\[
\|\mathbf{v}\| = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}
\]

### Step 3: Substitute into the cosine formula
Now substitute the values into the formula for \( \cos \theta \):

\[
\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}
\]

\[
\cos \theta = \frac{-1 + 4\sqrt{7}}{\sqrt{8} \cdot \sqrt{17}}
\]

First, calculate the denominator:

\[
\sqrt{8} \cdot \sqrt{17} = \sqrt{136}
\]

Thus:

\[
\cos \theta = \frac{-1 + 4\sqrt{7}}{\sqrt{136}}
\]

### Step 4: Solve for \( \theta \)
Using a calculator:

- \( 4\sqrt{7} \approx 10.583 \)
- \( -1 + 10.583 = 9.583 \)
- \( \sqrt{136} \approx 11.662 \)

Now:

\[
\cos \theta = \frac{9.583}{11.662} \approx 0.822
\]

Take the arccosine to find \( \theta \):

\[
\theta = \arccos(0.822)
\]

Using a calculator:

\[
\theta \approx 34.7^\circ
\]

### Final Answer:
The angle between the vectors is **34.7°**.

### Answer: D) 34.7°

---

Would you like further clarification or details on any of the steps? Here are some related questions:

1. What is the dot product of two vectors and why is it useful?
2. How do you calculate the magnitude of a vector in two dimensions?
3. How does the cosine formula help find the angle between vectors?
4. Can you explain the relationship between the dot product and angle?
5. What happens if the vectors are perpendicular or parallel?

**Tip:** To quickly identify perpendicular vectors, check if their dot product equals zero.

Find the angle between the given vectors to the nearest tenth of a degree.
\( \mathbf{u} = \mathbf{i} + \sqrt{7} \mathbf{j}, \mathbf{v} = -\mathbf{i} + 4\mathbf{j} \)
Options:
A) 22.2°   B) 69.5°   C) 6.7°   D) 34.7°

Discover how to calculate the angle between two vectors u = i + √7j and v = -i + 4j using vector algebra. This problem explores the dot product formula, vector magnitudes, and cosine relationships, providing a step-by-step solution for Grades 11-12 students.

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