If $(\vec{a} imes\vec{b})^{2}+(\vec{a}.\vec{b})^{2}=144$ and $|\vec{a}|=4,$ then $|\vec{b}|=$

Question

If $(\vec{a}	imes\vec{b})^{2}+(\vec{a}.\vec{b})^{2}=144$ and $|\vec{a}|=4,$ then $|\vec{b}|=$

Accepted Answer

To solve the problem, we need to understand the relationship between the given vectors $\vec{a}$ and $\vec{b}$. We are given:

$$
(\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2 = 144
$$

and

$$
|\vec{a}| = 4
$$

We need to find $|\vec{b}|$.

### Explanation:

1. **Vector Magnitude and Dot Product:**
   - The dot product $\vec{a} \cdot \vec{b}$ is given by $|\vec{a}||\vec{b}|\cos\theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

2. **Cross Product Magnitude:**
   - The magnitude of the cross product $|\vec{a} \times \vec{b}|$ is $|\vec{a}||\vec{b}|\sin\theta$.

3. **Substitute into the Given Equation:**
   - Substitute the expressions for the dot product and cross product into the given equation:
     $$
     (|\vec{a}||\vec{b}|\sin\theta)^2 + (|\vec{a}||\vec{b}|\cos\theta)^2 = 144
     $$

4. **Simplify:**
   - Factor out $(|\vec{a}||\vec{b}|)^2$:
     $$
     (|\vec{a}||\vec{b}|)^2(\sin^2\theta + \cos^2\theta) = 144
     $$

5. **Use the Pythagorean Identity:**
   - Since $\sin^2\theta + \cos^2\theta = 1$, the equation simplifies to:
     $$
     (|\vec{a}||\vec{b}|)^2 = 144
     $$

6. **Solve for $|\vec{b}|$:**
   - Substitute $|\vec{a}| = 4$:
     $$
     (4|\vec{b}|)^2 = 144
     $$
     $$
     16|\vec{b}|^2 = 144
     $$
     $$
     |\vec{b}|^2 = 9
     $$
     $$
     |\vec{b}| = 3
     $$

Thus, the magnitude of $|\vec{b}|$ is 3, which corresponds to Option 3.

### Explanation for Incorrect Options:

- **Option 1 (16):** If $|\vec{b}| = 16$, then $(4 \times 16)^2 = 4096$, which does not satisfy the equation $16|\vec{b}|^2 = 144$.

- **Option 2 (8):** If $|\vec{b}| = 8$, then $(4 \times 8)^2 = 1024$, which does not satisfy the equation $16|\vec{b}|^2 = 144$.

- **Option 4 (12):** If $|\vec{b}| = 12$, then $(4 \times 12)^2 = 2304$, which does not satisfy the equation $16|\vec{b}|^2 = 144$.

Therefore, Option 3 is the only correct answer, as it satisfies the given equation.