Two vector `vec(A)` and `vec(B)` have magnitudes A=3.00 and B=3.00. Their vector product is `vec(A)xxvec(B)=-5.00hat(k)+2.00hat(i)`.
What is the angle between `vec(A)` and `vec(B)`?

Explanation

To find the angle between the two vectors \(\vec{A}\) and \(\vec{B}\), we can follow these steps:

### Step 1: Understand the formula for the vector product
The magnitude of the vector product (cross product) of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]
where \(\theta\) is the angle between the two vectors.

### Step 2: Calculate the magnitude of the vector product
We are given the vector product:
\[ \vec{A} \times \vec{B} = -5 \hat{k} + 2 \hat{i} \]
To find its magnitude, we calculate:
\[ |\vec{A} \times \vec{B}| = \sqrt{(2)^2 + (0)^2 + (-5)^2} = \sqrt{4 + 0 + 25} = \sqrt{29} \]

### Step 3: Substitute known values into the formula
We know that the magnitudes of the vectors are:
\[ |\vec{A}| = 3.00 \quad \text{and} \quad |\vec{B}| = 3.00 \]
Substituting these values into the formula gives:
\[ \sqrt{29} = 3 \cdot 3 \cdot \sin \theta \]
This simplifies to:
\[ \sqrt{29} = 9 \sin \theta \]

### Step 4: Solve for \(\sin \theta\)
Rearranging the equation to isolate \(\sin \theta\):
\[ \sin \theta = \frac{\sqrt{29}}{9} \]

### Step 5: Calculate \(\theta\)
To find the angle \(\theta\), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{\sqrt{29}}{9}\right) \]

### Step 6: Compute the value
Using a calculator, we find:
\[ \theta \approx 36.752^\circ \]

Thus, the angle between the vectors \(\vec{A}\) and \(\vec{B}\) is approximately \(36.752^\circ\).

---