Key Concepts and Formulas

• Magnitude of Cross Product: For vectors $\vec{u}$ and $\vec{v}$ with angle $\theta$ between them, $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta$.
• Orthogonality of Cross Product: The cross product $\vec{u} \times \vec{v}$ is orthogonal to both $\vec{u}$ and $\vec{v}$, implying $(\vec{u} \times \vec{v}) \cdot \vec{u} = 0$ and $(\vec{u} \times \vec{v}) \cdot \vec{v} = 0$.
• Magnitude of Vector Difference Squared: $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.
• Trigonometric Optimization: Finding the minimum/maximum of a function over a given interval by analyzing its behavior.

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Step-by-Step Solution

Step 1: Utilize the orthogonality property derived from the cross product.

We are given $\vec{a} = \vec{b} \times \vec{c}$. A fundamental property of the cross product is that the resulting vector is orthogonal to both of the original vectors.

• Why this step is taken: This orthogonality simplifies the dot product terms that will appear when expanding $|\vec{c}-\vec{a}|^2$.
• Applying the concept: Since $\vec{a}$ is the cross product of $\vec{b}$ and $\vec{c}$, $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$. This means their dot products are zero: $\vec{a} \cdot \vec{b} = 0$ $\vec{a} \cdot \vec{c} = 0$

Step 2: Expand and simplify the term $|\vec{c}-\vec{a}|^2$.

We need to minimize $27|\vec{c}-\vec{a}|^2$. Let's first focus on the term $|\vec{c}-\vec{a}|^2$.

• Why this step is taken: Expanding the square of the vector difference allows us to use the dot product and incorporate the given magnitudes and the orthogonality established in Step 1.
• Applying the formula: Using the formula for the square of the magnitude of a vector difference: $|\vec{c}-\vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a})$
• Substitute known values: We are given $|\vec{a}|=2$. From Step 1, we know $\vec{c} \cdot \vec{a} = 0$. Substituting these values: $|\vec{c}-\vec{a}|^2 = |\vec{c}|^2 + (2)^2 - 2(0)$ $|\vec{c}-\vec{a}|^2 = |\vec{c}|^2 + 4$ Now, the problem reduces to finding the minimum value of $27(|\vec{c}|^2 + 4)$, which means we need to find the minimum value of $|\vec{c}|^2$.

Step 3: Express $|\vec{c}|$ in terms of the given angle $\alpha$ using the magnitude of the cross product.

The angle $\alpha$ is given as the angle between $\vec{b}$ and $\vec{c}$. The magnitude of the cross product relates these quantities.

• Why this step is taken: This step connects the unknown $|\vec{c}|$ to the given angle $\alpha$, which is constrained to an interval, allowing for optimization.
• Applying the formula: We take the magnitude of the given equation $\vec{a} = \vec{b} \times \vec{c}$: $|\vec{a}| = |\vec{b} \times \vec{c}|$ Using the magnitude of the cross product formula, $|\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\alpha$: $|\vec{a}| = |\vec{b}||\vec{c}|\sin\alpha$
• Substitute given values: We are given $|\vec{a}|=2$ and $|\vec{b}|=3$. $2 = 3|\vec{c}|\sin\alpha$
• Isolate $|\vec{c}|$: $|\vec{c}| = \frac{2}{3\sin\alpha}$
• Find $|\vec{c}|^2$: Squaring both sides: $|\vec{c}|^2 = \left(\frac{2}{3\sin\alpha}\right)^2 = \frac{4}{9\sin^2\alpha}$

Step 4: Substitute $|\vec{c}|^2$ back into the expression for $|\vec{c}-\vec{a}|^2$ and then find the minimum value of the target expression.

Now we have $|\vec{c}|^2$ in terms of $\alpha$, which we can substitute back into the expression from Step 2.

• Why this step is taken: This combines all the derived information to get the expression we need to minimize solely as a function of $\alpha$.
• Substitution: Substitute $|\vec{c}|^2 = \frac{4}{9\sin^2\alpha}$ into $|\vec{c}-\vec{a}|^2 = |\vec{c}|^2 + 4$: $|\vec{c}-\vec{a}|^2 = \frac{4}{9\sin^2\alpha} + 4$
• Target Expression: We need to minimize $27|\vec{c}-\vec{a}|^2$: $27|\vec{c}-\vec{a}|^2 = 27\left(\frac{4}{9\sin^2\alpha} + 4\right)$ $27|\vec{c}-\vec{a}|^2 = \frac{27 \cdot 4}{9\sin^2\alpha} + 27 \cdot 4$ $27|\vec{c}-\vec{a}|^2 = \frac{12}{\sin^2\alpha} + 108$

Step 5: Optimize the expression by considering the given interval for $\alpha$.

We need to find the minimum value of $\frac{12}{\sin^2\alpha} + 108$ for $\alpha \in \left[0, \frac{\pi}{3}\right]$.

• Why this step is taken: The expression to be minimized depends on $\sin^2\alpha$. We need to find the range of $\sin^2\alpha$ for the given interval of $\alpha$ to determine when the expression is minimized.
• Analyzing $\sin\alpha$ and $\sin^2\alpha$: For $\alpha \in \left[0, \frac{\pi}{3}\right]$: The function $\sin\alpha$ is increasing in this interval. The minimum value of $\sin\alpha$ is $\sin(0) = 0$. The maximum value of $\sin\alpha$ is $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$. Therefore, $\sin\alpha \in \left[0, \frac{\sqrt{3}}{2}\right]$. Now consider $\sin^2\alpha$. Since $\sin\alpha$ can be 0, $\sin^2\alpha$ can also be 0. However, if $\sin\alpha = 0$, the expression $\frac{12}{\sin^2\alpha}$ is undefined. We must consider the domain of $\alpha$ carefully. The problem states $\alpha$ is the angle between $\vec{b}$ and $\vec{c}$. If $\sin\alpha = 0$, then $\alpha = 0$ or $\alpha = \pi$. If $\alpha=0$, $\vec{b}$ and $\vec{c}$ are parallel, and their cross product $\vec{a}$ would be the zero vector, $|\vec{a}|=0$, contradicting $|\vec{a}|=2$. Thus, $\sin\alpha \neq 0$. So, for $\alpha \in \left(0, \frac{\pi}{3}\right]$, $\sin\alpha \in \left(0, \frac{\sqrt{3}}{2}\right]$. Squaring these values, $\sin^2\alpha \in \left(0, \left(\frac{\sqrt{3}}{2}\right)^2\right] = \left(0, \frac{3}{4}\right]$.
• Minimizing the expression: The expression is $\frac{12}{\sin^2\alpha} + 108$. To minimize this expression, we need to maximize the denominator, $\sin^2\alpha$. The maximum value of $\sin^2\alpha$ in the interval $\left(0, \frac{3}{4}\right]$ is $\frac{3}{4}$, which occurs when $\alpha = \frac{\pi}{3}$.
• Calculate the minimum value: Substitute the maximum value of $\sin^2\alpha = \frac{3}{4}$ into the expression: Minimum value $= \frac{12}{3/4} + 108$ Minimum value $= 12 \cdot \frac{4}{3} + 108$ Minimum value $= 4 \cdot 4 + 108$ Minimum value $= 16 + 108$ Minimum value $= 124$

Common Mistakes & Tips

• Division by Zero: Be careful when $\sin\alpha = 0$. In this problem, $\sin\alpha=0$ implies $|\vec{a}|=0$, which contradicts the given $|\vec{a}|=2$. Thus, $\alpha \neq 0$.
• Interval Analysis: Ensure you correctly determine the range of $\sin^2\alpha$ for the given interval of $\alpha$. The function $\sin^2\alpha$ is not monotonic on $[0, \pi/3]$ if the interval included $\pi/2$ or beyond.
• Orthogonality: Always remember that $\vec{a} = \vec{b} \times \vec{c}$ implies $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$. This is a key simplification.

Summary

The problem required us to minimize $27|\vec{c}-\vec{a}|^2$. We first simplified $|\vec{c}-\vec{a}|^2$ using the property that $\vec{a}$ is orthogonal to $\vec{c}$, reducing it to $|\vec{c}|^2 + 4$. Then, we used the magnitude of the cross product formula, $|\vec{a}| = |\vec{b}||\vec{c}|\sin\alpha$, to express $|\vec{c}|^2$ in terms of $\sin^2\alpha$: $|\vec{c}|^2 = \frac{4}{9\sin^2\alpha}$. Substituting this into the expression to be minimized gave us $\frac{12}{\sin^2\alpha} + 108$. Finally, by analyzing the range of $\sin^2\alpha$ for $\alpha \in \left[0, \frac{\pi}{3}\right]$ (specifically $\sin^2\alpha \in \left(0, \frac{3}{4}\right]$), we found that the expression is minimized when $\sin^2\alpha$ is maximized at $\frac{3}{4}$, leading to a minimum value of 124.

The final answer is \boxed{124}.