If `|vec(a)|=10,|vec(b)|=2` and `vec(a)*vec(b)=12`, then the value of the `|vec(a)xxvec(b)|` 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:

1. \(|\vec{a}| = 10\)
2. \(|\vec{b}| = 2\)
3. \(\vec{a} \cdot \vec{b} = 12\)

### Step-by-Step Solution:

**Step 1: Use the dot product formula.**

The dot product of two vectors can be expressed as:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
where \(\theta\) is the angle between the two vectors.

**Step 2: Substitute the known values into the dot product formula.**

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

Substituting these values into the formula gives:
\[ 12 = 10 \cdot 2 \cdot \cos \theta \]
This simplifies to:
\[ 12 = 20 \cos \theta \]

**Step 3: Solve for \(\cos \theta\).**

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

**Step 4: Use the Pythagorean identity to find \(\sin \theta\).**

We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \(\cos \theta = \frac{3}{5}\):
\[ \sin^2 \theta + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 \theta + \frac{9}{25} = 1 \]
\[ \sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \]
Taking the square root gives:
\[ \sin \theta = \frac{4}{5} \]

**Step 5: Use the cross product formula to find \(|\vec{a} \times \vec{b}|\).**

The magnitude of the cross product is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
Substituting the known values:
\[ |\vec{a} \times \vec{b}| = 10 \cdot 2 \cdot \frac{4}{5} \]
Calculating this gives:
\[ |\vec{a} \times \vec{b}| = 20 \cdot \frac{4}{5} = 16 \]

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

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