Step-by-step Solution:

Correct Answer: B (3)

The value of $\lambda$ can be found by using the given magnitude of vector $\vec{b}$ and the properties of the dot product.

Step 1: Set up the Magnitude Equation

We are given the magnitude of vector $\vec{b}$, which is $|\vec{b}| = 4$. The square of the magnitude of a vector is equal to the dot product of the vector with itself, so $|\vec{b}|^2 = \vec{b} \cdot \vec{b} = 16$.

We are also given the expression for $\vec{b} = 2\vec{c} + \lambda\vec{a}$. Let's substitute this into the equation:

$$ |\vec{b}|^2 = (2\vec{c} + \lambda\vec{a}) \cdot (2\vec{c} + \lambda\vec{a}) = 16 $$

Step 2: Expand and Simplify the Dot Product

Using the distributive property of the dot product, we can expand the expression:

$$ (2\vec{c} \cdot 2\vec{c}) + (2\vec{c} \cdot \lambda\vec{a}) + (\lambda\vec{a} \cdot 2\vec{c}) + (\lambda\vec{a} \cdot \lambda\vec{a}) = 16 $$

$$ 4(\vec{c} \cdot \vec{c}) + 2\lambda(\vec{c} \cdot \vec{a}) + 2\lambda(\vec{a} \cdot \vec{c}) + \lambda^2(\vec{a} \cdot \vec{a}) = 16 $$

Using the properties $\vec{v} \cdot \vec{v} = |\vec{v}|^2$ and $\vec{a} \cdot \vec{c} = \vec{c} \cdot \vec{a}$, we get:

$$ 4|\vec{c}|^2 + 4\lambda(\vec{a} \cdot \vec{c}) + \lambda^2|\vec{a}|^2 = 16 $$

Step 3: Substitute the Given Values

The problem provides the following information:

• $\vec{a}$ and $\vec{c}$ are unit vectors, so $|\vec{a}| = 1$ and $|\vec{c}| = 1$.
• The angle $\theta$ between $\vec{a}$ and $\vec{c}$ is $\cos^{-1}(\frac{1}{4})$, which means $\cos\theta = \frac{1}{4}$.
• The dot product is $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos\theta = (1)(1)(\frac{1}{4}) = \frac{1}{4}$.

Now, substitute these values into our simplified equation:

$$ 4(1)^2 + 4\lambda\left(\frac{1}{4}\right) + \lambda^2(1)^2 = 16 $$

$$ 4 + \lambda + \lambda^2 = 16 $$

Step 4: Solve for $\lambda$

Rearrange the terms to form a standard quadratic equation:

$$ \lambda^2 + \lambda - 12 = 0 $$

This equation can be factored:

$$ (\lambda + 4)(\lambda - 3) = 0 $$

The two possible solutions are $\lambda = -4$ and $\lambda = 3$.

Since the problem states that $\lambda > 0$, we choose the positive solution.

$$ \lambda = 3 $$