To find the angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$, we use the dot product formula:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

Where:

• $\vec{a} \cdot \vec{b}$ is the dot product of $\vec{a}$ and $\vec{b}$,
• $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of $\vec{a}$ and $\vec{b}$, respectively,
• $\theta$ is the angle between the vectors.

Step 1: Compute the dot product $\vec{a} \cdot \vec{b}$

Given vectors: $\vec{a} = \sqrt{3}\hat{i} - 2\hat{j} + 5\hat{k}$ $\vec{b} = 2\hat{i} + \sqrt{3}\hat{j} + \hat{k}$

The dot product $\vec{a} \cdot \vec{b}$ is computed as: $\vec{a} \cdot \vec{b} = (\sqrt{3})(2) + (-2)(\sqrt{3}) + (5)(1)$ $\vec{a} \cdot \vec{b} = 2\sqrt{3} - 2\sqrt{3} + 5 = 5$

Step 2: Compute the magnitudes $|\vec{a}|$ and $|\vec{b}|$

Magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{(\sqrt{3})^2 + (-2)^2 + 5^2}$ $|\vec{a}| = \sqrt{3 + 4 + 25} = \sqrt{32} = 4\sqrt{2}$

Magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{(2)^2 + (\sqrt{3})^2 + (1)^2}$ $|\vec{b}| = \sqrt{4 + 3 + 1} = \sqrt{8} = 2\sqrt{2}$

Step 3: Compute the cosine of the angle

Now, use the dot product formula to find $\cos \theta$: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{5}{(4\sqrt{2})(2\sqrt{2})} = \frac{5}{16}$

Step 4: Find the angle $\theta$

$\theta = \cos^{-1}\left(\frac{5}{16}\right)$

So, the angle between the vectors is:

$\theta = \cos^{-1}\left(\frac{5}{16}\right) \approx 72.54^\circ$

Would you like more details or have any questions?

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Here are 5 related questions:

1. How can the cross product of two vectors be used to find the area of a parallelogram?
2. What are the applications of vector dot product in physics?
3. How do you find the projection of one vector onto another?
4. Can the angle between two vectors ever be greater than 180°? Why or why not?
5. How does the dot product change if one of the vectors is multiplied by a scalar?

Tip: The dot product of orthogonal vectors (perpendicular) is always zero because $\cos(90^\circ) = 0$.