If `|vec(a) | = 3, | vec(b) | =4` and `| vec(a) xx vec(b) | = 10`, then `| vec(a). vec(b) |^(2)` is equal to

Options

A. 22
B. 88
C. 44
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given:
- \( |\vec{a}| = 3 \)
- \( |\vec{b}| = 4 \)
- \( |\vec{a} \times \vec{b}| = 10 \)

### Step 2: Use the Formula for the Magnitude of the Cross Product
The magnitude of the cross product of two vectors can be expressed as:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
where \( \theta \) is the angle between the two vectors.

### Step 3: Substitute the Known Values into the Formula
Substituting the known values into the formula:
\[ 10 = 3 \cdot 4 \cdot \sin \theta \]
This simplifies to:
\[ 10 = 12 \sin \theta \]

### Step 4: Solve for \( \sin \theta \)
To find \( \sin \theta \), we rearrange the equation:
\[ \sin \theta = \frac{10}{12} = \frac{5}{6} \]

### Step 5: Use the Pythagorean Identity to Find \( \cos \theta \)
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \( \sin \theta \):
\[ \left(\frac{5}{6}\right)^2 + \cos^2 \theta = 1 \]
Calculating \( \left(\frac{5}{6}\right)^2 \):
\[ \frac{25}{36} + \cos^2 \theta = 1 \]
Rearranging gives:
\[ \cos^2 \theta = 1 - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36} \]

### Step 6: Use the Dot Product Formula
The dot product of two vectors can be expressed as:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the known values:
\[ \vec{a} \cdot \vec{b} = 3 \cdot 4 \cdot \cos \theta \]
We know \( \cos \theta = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6} \):
\[ \vec{a} \cdot \vec{b} = 12 \cdot \frac{\sqrt{11}}{6} = 2\sqrt{11} \]

### Step 7: Find \( |\vec{a} \cdot \vec{b}|^2 \)
Now we need to find \( |\vec{a} \cdot \vec{b}|^2 \):
\[ |\vec{a} \cdot \vec{b}|^2 = (2\sqrt{11})^2 = 4 \cdot 11 = 44 \]

### Final Answer
Thus, the value of \( |\vec{a} \cdot \vec{b}|^2 \) is \( \boxed{44} \).