The resultant of the two vectors having magnitude 2 and 3 is 1. What is their cross product

Options

A. 6
B. 3
C. 1
D. 0

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the given information
We are given two vectors \( \vec{A} \) and \( \vec{B} \) with magnitudes:
- \( |\vec{A}| = 2 \)
- \( |\vec{B}| = 3 \)

The resultant of these two vectors is given as:
- \( |\vec{A} + \vec{B}| = 1 \)

### Step 2: Use the formula for the magnitude of the resultant
The magnitude of the resultant of two vectors can be expressed using the formula:
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
where \( \theta \) is the angle between the two vectors.

### Step 3: Substitute the known values into the formula
Substituting the values we have:
\[ 1 = \sqrt{2^2 + 3^2 + 2 \cdot 2 \cdot 3 \cdot \cos \theta} \]
This simplifies to:
\[ 1 = \sqrt{4 + 9 + 12 \cos \theta} \]
\[ 1 = \sqrt{13 + 12 \cos \theta} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 1^2 = 13 + 12 \cos \theta \]
\[ 1 = 13 + 12 \cos \theta \]

### Step 5: Solve for \( \cos \theta \)
Rearranging the equation:
\[ 12 \cos \theta = 1 - 13 \]
\[ 12 \cos \theta = -12 \]
\[ \cos \theta = -1 \]

### Step 6: Determine the angle \( \theta \)
The angle \( \theta \) for which \( \cos \theta = -1 \) is:
\[ \theta = 180^\circ \text{ or } \pi \text{ radians} \]

### Step 7: Calculate the cross product
The cross product \( \vec{A} \times \vec{B} \) is given by:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]
Substituting the values:
\[ |\vec{A} \times \vec{B}| = 2 \cdot 3 \cdot \sin(180^\circ) \]
Since \( \sin(180^\circ) = 0 \):
\[ |\vec{A} \times \vec{B}| = 6 \cdot 0 = 0 \]

### Final Answer
The cross product \( \vec{A} \times \vec{B} \) is \( 0 \).

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