Which of the following is the unit vector perpendicular to `vec(A)` and `vec(B)`?

Options

A. `(hatAxxhatB)/(ABsintheta)`
B. `(hatAxxhatB)/(ABcostheta)`
C. `(AxxB)/(ABsintheta)`
D. `(AxxB)/(ABcostheta)`

Answer

To find a unit vector that is perpendicular to two vectors \(\vec{A}\) and \(\vec{B}\), we can follow these steps:

### Step 1: Understand the Cross Product
The cross product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \vec{A} \times \vec{B} \]
This operation yields a vector that is perpendicular to both \(\vec{A}\) and \(\vec{B}\).

### Step 2: Calculate the Magnitude of the Cross Product
The magnitude of the cross product \(\vec{A} \times \vec{B}\) is given by:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]
where \(\theta\) is the angle between the two vectors.

### Step 3: Normalize the Cross Product
To convert the cross product into a unit vector, we need to divide by its magnitude:
\[ \text{Unit vector} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} \]
Substituting the magnitude from Step 2, we get:
\[ \text{Unit vector} = \frac{\vec{A} \times \vec{B}}{|\vec{A}| |\vec{B}| \sin \theta} \]

### Step 4: Identify the Correct Option
From the options given, we need to find the one that matches the expression derived in Step 3. The correct unit vector perpendicular to both \(\vec{A}\) and \(\vec{B}\) is:
\[ \frac{\vec{A} \times \vec{B}}{|\vec{A}| |\vec{B}| \sin \theta} \]
This matches with the option that states the unit vector is derived from the cross product divided by the appropriate magnitude.

### Conclusion
Thus, the unit vector perpendicular to \(\vec{A}\) and \(\vec{B}\) is:
\[ \frac{\vec{A} \times \vec{B}}{|\vec{A}| |\vec{B}| \sin \theta} \]

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