Key Concepts and Formulas

• The angle $\theta$ between two non-zero vectors $\vec{x}$ and $\vec{y}$ is given by $\cos \theta = \frac{\vec{x} \cdot \vec{y}}{|\vec{x}| |\vec{y}|}$.
• The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is zero if any two of the vectors are identical or parallel.
• The dot product of a vector with itself is the square of its magnitude: $\vec{v} \cdot \vec{v} = |\vec{v}|^2$.
• Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$.

Step-by-Step Solution

We are given vectors $\vec{a}$ and $\vec{b}$ with $|\vec{a}|=1$, $|\vec{b}|=4$, and $\vec{a} \cdot \vec{b}=2$. We are also given $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$. We need to find the angle between $\vec{b}$ and $\vec{c}$.

Step 1: Calculate the dot product $\vec{b} \cdot \vec{c}$

To find the angle between $\vec{b}$ and $\vec{c}$, we first need to compute their dot product: $\vec{b} \cdot \vec{c} = \vec{b} \cdot (2(\vec{a} \times \vec{b})-3 \vec{b})$ Using the distributive property of the dot product: $\vec{b} \cdot \vec{c} = 2[\vec{b} \cdot (\vec{a} \times \vec{b})] - 3(\vec{b} \cdot \vec{b})$ The term $\vec{b} \cdot (\vec{a} \times \vec{b})$ is a scalar triple product where the vector $\vec{b}$ is repeated. A key property of the scalar triple product is that it is zero if any two vectors are identical or parallel. Thus, $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$. The term $\vec{b} \cdot \vec{b}$ is the dot product of a vector with itself, which equals the square of its magnitude: $\vec{b} \cdot \vec{b} = |\vec{b}|^2$. We are given $|\vec{b}|=4$, so $|\vec{b}|^2 = 4^2 = 16$. Substituting these values back: $\vec{b} \cdot \vec{c} = 2(0) - 3(16)$ $\vec{b} \cdot \vec{c} = -48$

Step 2: Calculate the magnitude of $\vec{c}$, i.e., $|\vec{c}|$

We will find $|\vec{c}|^2$ first, using $|\vec{c}|^2 = \vec{c} \cdot \vec{c}$. $|\vec{c}|^2 = (2(\vec{a} \times \vec{b})-3 \vec{b}) \cdot (2(\vec{a} \times \vec{b})-3 \vec{b})$ Expanding this expression: $|\vec{c}|^2 = (2(\vec{a} \times \vec{b})) \cdot (2(\vec{a} \times \vec{b})) - 2[(2(\vec{a} \times \vec{b})) \cdot (3 \vec{b})] + (3 \vec{b}) \cdot (3 \vec{b})$ $|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 - 12[(\vec{a} \times \vec{b}) \cdot \vec{b}] + 9|\vec{b}|^2$ As established in Step 1, the term $(\vec{a} \times \vec{b}) \cdot \vec{b}$ is a scalar triple product with a repeated vector, so it is zero. $|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 - 12(0) + 9|\vec{b}|^2$ $|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2$ Now we need to find $|\vec{a} \times \vec{b}|^2$. We use Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$. Substituting the given values $|\vec{a}|=1$, $|\vec{b}|=4$, and $\vec{a} \cdot \vec{b}=2$: $|\vec{a} \times \vec{b}|^2 = (1)^2 (4)^2 - (2)^2$ $|\vec{a} \times \vec{b}|^2 = 1 \times 16 - 4$ $|\vec{a} \times \vec{b}|^2 = 16 - 4 = 12$ Now substitute this and $|\vec{b}|^2 = 16$ back into the expression for $|\vec{c}|^2$: $|\vec{c}|^2 = 4(12) + 9(16)$ $|\vec{c}|^2 = 48 + 144$ $|\vec{c}|^2 = 192$ Taking the square root to find $|\vec{c}|$: $|\vec{c}| = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$

Step 3: Determine the angle $\theta$ between $\vec{b}$ and $\vec{c}$

Using the formula for the angle between two vectors: $\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$ Substitute the calculated values: $\vec{b} \cdot \vec{c} = -48$, $|\vec{b}|=4$, and $|\vec{c}|=8\sqrt{3}$: $\cos \theta = \frac{-48}{(4)(8\sqrt{3})}$ $\cos \theta = \frac{-48}{32\sqrt{3}}$ Simplify the fraction: $\cos \theta = \frac{-3}{2\sqrt{3}}$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $\cos \theta = \frac{-3\sqrt{3}}{2\sqrt{3} \cdot \sqrt{3}} = \frac{-3\sqrt{3}}{2 \cdot 3} = \frac{-3\sqrt{3}}{6}$ $\cos \theta = -\frac{\sqrt{3}}{2}$ Therefore, the angle $\theta$ is: $\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

Common Mistakes & Tips

• Scalar Triple Product: Always remember that if a scalar triple product involves a repeated vector, its value is zero. This is a critical shortcut.
• Lagrange's Identity: This identity is essential for problems involving the magnitude of a cross product when dot products and individual magnitudes are known.
• Algebraic Simplification: Pay close attention to signs and simplifying radicals. Rationalizing the denominator is standard practice.

Summary

We calculated the dot product $\vec{b} \cdot \vec{c}$ by utilizing the distributive property of the dot product and the property of the scalar triple product. We then found the magnitude of $\vec{c}$ by first calculating $|\vec{c}|^2$ and employing Lagrange's identity to find $|\vec{a} \times \vec{b}|^2$. Finally, we substituted these values into the cosine formula for the angle between two vectors to arrive at the result.

The final answer is $\boxed{\text{(D)}}$.