If `|veca| = 10, |vecb|=2` and `veca.vecb=12`, then the value of `|veca xx vecb|` is:

Options

A. 5
B. 10
C. 14
D. 16

Answer

To solve the problem, we need to find the magnitude of the cross product of two vectors \(\vec{a}\) and \(\vec{b}\) given the following information:

- \(|\vec{a}| = 10\)
- \(|\vec{b}| = 2\)
- \(\vec{a} \cdot \vec{b} = 12\)

### Step-by-Step Solution:

1. **Use the Dot Product Formula**:
The dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
where \(\theta\) is the angle between the two vectors.

2. **Substitute the Known Values**:
We know:
\[ \vec{a} \cdot \vec{b} = 12, \quad |\vec{a}| = 10, \quad |\vec{b}| = 2 \]
Substituting these values into the dot product formula:
\[ 12 = 10 \cdot 2 \cdot \cos \theta \]
Simplifying gives:
\[ 12 = 20 \cos \theta \]

3. **Solve for \(\cos \theta\)**:
Rearranging the equation to find \(\cos \theta\):
\[ \cos \theta = \frac{12}{20} = \frac{3}{5} \]

4. **Use the Pythagorean Identity**:
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \(\cos^2 \theta\):
\[ \sin^2 \theta + \left(\frac{3}{5}\right)^2 = 1 \]
This simplifies to:
\[ \sin^2 \theta + \frac{9}{25} = 1 \]
Therefore:
\[ \sin^2 \theta = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25} \]

5. **Calculate \(\sin \theta\)**:
Taking the square root:
\[ \sin \theta = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

6. **Calculate the Magnitude of the Cross Product**:
The magnitude of the cross product \(|\vec{a} \times \vec{b}|\) is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
Substituting the known values:
\[ |\vec{a} \times \vec{b}| = 10 \cdot 2 \cdot \frac{4}{5} \]
Simplifying:
\[ |\vec{a} \times \vec{b}| = 20 \cdot \frac{4}{5} = 16 \]

### Final Answer:
Thus, the value of \(|\vec{a} \times \vec{b}|\) is \(16\).