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.