We are given two vectors:

$\vec{a} = 6.0 \hat{i} + 6.0 \hat{j} + 6.0 \hat{k}$ $\vec{b} = 3.0 \hat{i} + 5.0 \hat{j} + 6.0 \hat{k}$

To calculate the angle $\theta$ between the two vectors, we will use the scalar (dot) product formula:

$\vec{a} \cdot \vec{b} = ab \cos \theta$ where $a$ and $b$ are the magnitudes of the vectors $\vec{a}$ and $\vec{b}$, and $\theta$ is the angle between them.

1. Calculate the dot product $\vec{a} \cdot \vec{b}$:

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

Substitute the components of the vectors $\vec{a}$ and $\vec{b}$:

$\vec{a} \cdot \vec{b} = (6.0)(3.0) + (6.0)(5.0) + (6.0)(6.0)$ $\vec{a} \cdot \vec{b} = 18.0 + 30.0 + 36.0 = 84.0$

2. Calculate the magnitudes of $\vec{a}$ and $\vec{b}$:

$|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{(6.0)^2 + (6.0)^2 + (6.0)^2} = \sqrt{108} \approx 10.39$ $|\vec{b}| = \sqrt{b_x^2 + b_y^2 + b_z^2} = \sqrt{(3.0)^2 + (5.0)^2 + (6.0)^2} = \sqrt{70} \approx 8.37$

3. Use the dot product formula to solve for $\cos \theta$:

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{84.0}{10.39 \times 8.37} \approx \frac{84.0}{87.01} \approx 0.966$

4. Find the angle $\theta$:

$\theta = \cos^{-1}(0.966) \approx 15.98^\circ$

Thus, the angle between the two vectors is approximately 16 degrees.

Would you like more details on any part of this process?

Related Questions:

1. How can you calculate the cross product of two vectors?
2. What is the physical meaning of the dot product in terms of projection?
3. How does the dot product change if the vectors are perpendicular?
4. Can the dot product of two vectors ever be negative, and what does it signify?
5. How would the result change if one of the vector components was zero?

Tip:

When vectors are perpendicular, their dot product is zero, and the angle between them is $90^\circ$.