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The problem asks to calculate the angle between two vectors using the definition of the scalar (dot) product. The vectors provided are:

\[
\vec{a} = 3.0\hat{i} + 3.0\hat{j} + 3.0\hat{k}
\]
and
\[
\vec{b} = 8.0\hat{i} + 7.0\hat{j} + 7.0\hat{k}.
\]

### Step 1: Dot Product Formula

The scalar product (dot product) between two vectors \(\vec{a}\) and \(\vec{b}\) is given by:

\[
\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z
\]

Here,
\[
a_x = 3.0, \quad a_y = 3.0, \quad a_z = 3.0
\]
\[
b_x = 8.0, \quad b_y = 7.0, \quad b_z = 7.0
\]

Substitute the values into the dot product formula:

\[
\vec{a} \cdot \vec{b} = (3.0)(8.0) + (3.0)(7.0) + (3.0)(7.0)
\]

### Step 2: Magnitude of Vectors

The magnitude of a vector \(\vec{a}\) is calculated using:

\[
|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}
\]

For \(\vec{a} = 3.0\hat{i} + 3.0\hat{j} + 3.0\hat{k}\):

\[
|\vec{a}| = \sqrt{(3.0)^2 + (3.0)^2 + (3.0)^2} = \sqrt{9 + 9 + 9} = \sqrt{27}
\]

For \(\vec{b} = 8.0\hat{i} + 7.0\hat{j} + 7.0\hat{k}\):

\[
|\vec{b}| = \sqrt{(8.0)^2 + (7.0)^2 + (7.0)^2} = \sqrt{64 + 49 + 49} = \sqrt{162}
\]

### Step 3: Cosine of the Angle

The dot product is also related to the angle \(\theta\) between the vectors by the equation:

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

Therefore, the cosine of the angle is:

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

### Step 4: Calculate the Angle

Let me compute all the necessary values.Here are the results of the calculation:

- The dot product \(\vec{a} \cdot \vec{b} = 66.0\).
- The magnitude of \(\vec{a} = 5.196\).
- The magnitude of \(\vec{b} = 12.728\).
- The cosine of the angle \(\cos \theta = 0.998\).
- The angle \(\theta\) between the two vectors is approximately \(3.68^\circ\).

Would you like further clarification or more details?

Here are some related questions you might want to explore:
1. How is the dot product used in physics?
2. Can the angle between two vectors ever be greater than \(180^\circ\)?
3. How does the magnitude of vectors affect their direction?
4. What happens to the angle if one of the vectors is multiplied by a scalar?
5. How does the cross product differ from the dot product in vector operations?

**Tip:** When vectors are nearly parallel, their dot product is close to the product of their magnitudes, resulting in a small angle between them.

Use the definition of scalar product and the fact that a · b = a_x*b_x + a_y*b_y + a_z*b_z to calculate the angle between the two vectors a = 3.0i + 3.0j + 3.0k and b = 8.0i + 7.0j + 7.0k.

Learn how to calculate the angle between two vectors a = 3.0i + 3.0j + 3.0k and b = 8.0i + 7.0j + 7.0k using the dot product. This problem covers key concepts in vector algebra and geometry, including the dot product and magnitude of vectors. Suitable for students in Grades 11-12.

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