If `|vec(A)| = 4N, |vec(B)| = 3N` the value of `|vec(A) xx vec(B)|^(2) + |vec(A).vec(B)|^(2) =`

Options

A. `5N`
B. `25N`
C. `144N`
D. `169N`

Answer

To solve the problem, we need to find the value of \( |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 \) given that \( |\vec{A}| = 4 \, \text{N} \) and \( |\vec{B}| = 3 \, \text{N} \).

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

The magnitude of the cross product is given by:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]
The magnitude of the dot product is given by:
\[ |\vec{A} \cdot \vec{B}| = |\vec{A}| |\vec{B}| \cos \theta \]

### Step 2: Square both expressions.

Now, we will square both the cross product and the dot product:
\[ |\vec{A} \times \vec{B}|^2 = (|\vec{A}| |\vec{B}| \sin \theta)^2 = |\vec{A}|^2 |\vec{B}|^2 \sin^2 \theta \]
\[ |\vec{A} \cdot \vec{B}|^2 = (|\vec{A}| |\vec{B}| \cos \theta)^2 = |\vec{A}|^2 |\vec{B}|^2 \cos^2 \theta \]

### Step 3: Add the squared expressions.

Now we add the two squared expressions:
\[ |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = |\vec{A}|^2 |\vec{B}|^2 \sin^2 \theta + |\vec{A}|^2 |\vec{B}|^2 \cos^2 \theta \]
Factoring out \( |\vec{A}|^2 |\vec{B}|^2 \):
\[ = |\vec{A}|^2 |\vec{B}|^2 (\sin^2 \theta + \cos^2 \theta) \]

### Step 4: Use the Pythagorean identity.

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = |\vec{A}|^2 |\vec{B}|^2 \cdot 1 = |\vec{A}|^2 |\vec{B}|^2 \]

### Step 5: Substitute the magnitudes of vectors A and B.

Now substitute the given magnitudes:
\[ |\vec{A}| = 4 \, \text{N} \quad \text{and} \quad |\vec{B}| = 3 \, \text{N} \]
\[ |\vec{A}|^2 = 4^2 = 16 \quad \text{and} \quad |\vec{B}|^2 = 3^2 = 9 \]
Thus,
\[ |\vec{A}|^2 |\vec{B}|^2 = 16 \times 9 = 144 \]

### Final Answer

Therefore, the value of \( |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 \) is:
\[ \boxed{144} \]