The angle between vectors `vec(A) and vec(B)` is `60^@` What is the ratio `vec(A) .vec(B)` and `|vec(A) xxvec(B)|`

Options

A. `1/(sqrt(2))`
B. `1/(sqrt(3))`
C. `1/(sqrt(4))`
D. `1/(sqrt(5))`

Answer

To solve the problem of finding the ratio of the dot product and the magnitude of the cross product of two vectors \( \vec{A} \) and \( \vec{B} \) given that the angle between them is \( 60^\circ \), we can follow these steps:

### Step 1: Write the formulas for dot product and cross product
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 vectors.

The magnitude of the cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is given by:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin(\theta) \]

### Step 2: Substitute the angle into the formulas
Given that \( \theta = 60^\circ \):
- The cosine of \( 60^\circ \) is \( \cos(60^\circ) = \frac{1}{2} \).
- The sine of \( 60^\circ \) is \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \).

### Step 3: Write the expressions for dot product and cross product
Now substituting the values into the formulas:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cdot \frac{1}{2} \]
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \cdot \frac{\sqrt{3}}{2} \]

### Step 4: Find the ratio of the dot product to the magnitude of the cross product
To find the ratio \( \frac{\vec{A} \cdot \vec{B}}{|\vec{A} \times \vec{B}|} \):
\[ \frac{\vec{A} \cdot \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{|\vec{A}| |\vec{B}| \cdot \frac{1}{2}}{|\vec{A}| |\vec{B}| \cdot \frac{\sqrt{3}}{2}} \]

### Step 5: Simplify the ratio
The \( |\vec{A}| |\vec{B}| \) terms cancel out:
\[ \frac{\vec{A} \cdot \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, the ratio of \( \vec{A} \cdot \vec{B} \) to \( |\vec{A} \times \vec{B}| \) is:
\[ \frac{1}{\sqrt{3}} \]