Key Concepts and Formulas

• Unit Vectors: A vector $\vec{v}$ is a unit vector if $|\vec{v}| = 1$. Consequently, $|\vec{v}|^2 = \vec{v} \cdot \vec{v} = 1$.
• Dot Product: For vectors $\vec{x}$ and $\vec{y}$, $\vec{x} \cdot \vec{y} = |\vec{x}| |\vec{y}| \cos \theta$, where $\theta$ is the angle between them.
• Cross Product Magnitude: For vectors $\vec{x}$ and $\vec{y}$, $|\vec{x} \times \vec{y}| = |\vec{x}| |\vec{y}| \sin \theta$.
• Trigonometric Identity: $\sin^2 \theta + \cos^2 \theta = 1$.

Step-by-Step Solution

Step 1: Rearrange the Given Vector Equation We are given the equation $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$. To work with magnitudes and dot products, it's useful to isolate a term. Let's isolate the term involving $\vec{b}$ because its magnitude is known (as it's a unit vector), and we want to find $|\vec{a} \times \vec{c}|$. $\vec{a} + 2\vec{c} = -2\vec{b}$ Reasoning: This step prepares the equation for squaring both sides, which will allow us to eliminate $\vec{b}$ from the primary equation by using its known magnitude.

Step 2: Square Both Sides of the Equation Now, we square the magnitude of both sides of the rearranged equation. Recall that for any vector $\vec{v}$, $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$. $|\vec{a} + 2\vec{c}|^2 = |-2\vec{b}|^2$ Expanding the left side using the property $|\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$: $|\vec{a}|^2 + |2\vec{c}|^2 + 2(\vec{a} \cdot 2\vec{c}) = (-2)^2 |\vec{b}|^2$ $|\vec{a}|^2 + 4|\vec{c}|^2 + 4(\vec{a} \cdot \vec{c}) = 4|\vec{b}|^2$ Reasoning: Squaring both sides converts the vector sum into an equation involving magnitudes and dot products, which can be solved using the properties of vectors.

Step 3: Substitute Unit Vector Magnitudes We are given that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors. Therefore, their magnitudes are all 1: $|\vec{a}| = 1$, $|\vec{b}| = 1$, and $|\vec{c}| = 1$. Substitute these values into the equation from Step 2. $(1)^2 + 4(1)^2 + 4(\vec{a} \cdot \vec{c}) = 4(1)^2$ $1 + 4 + 4(\vec{a} \cdot \vec{c}) = 4$ $5 + 4(\vec{a} \cdot \vec{c}) = 4$ Reasoning: This step simplifies the equation significantly by using the given information about the vectors being unit vectors.

Step 4: Solve for the Dot Product $\vec{a} \cdot \vec{c}$ From the simplified equation in Step 3, we can now solve for the dot product $\vec{a} \cdot \vec{c}$. $4(\vec{a} \cdot \vec{c}) = 4 - 5$ $4(\vec{a} \cdot \vec{c}) = -1$ $\vec{a} \cdot \vec{c} = -\frac{1}{4}$ Reasoning: We have found the value of the dot product of $\vec{a}$ and $\vec{c}$. This value is directly related to the cosine of the angle between them.

Step 5: Determine $\cos \theta$ between $\vec{a}$ and $\vec{c}$ The definition of the dot product is $\vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{c}$. Since $\vec{a}$ and $\vec{c}$ are unit vectors, $|\vec{a}|=1$ and $|\vec{c}|=1$. $-\frac{1}{4} = (1)(1) \cos \theta$ $\cos \theta = -\frac{1}{4}$ Reasoning: This step explicitly finds the cosine of the angle between $\vec{a}$ and $\vec{c}$, which is a necessary component for calculating the magnitude of their cross product.

Step 6: Calculate $\sin \theta$ using the Trigonometric Identity To find $|\vec{a} \times \vec{c}|$, we need $\sin \theta$. We use the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$. $\sin^2 \theta = 1 - \cos^2 \theta$ Substitute the value of $\cos \theta$: $\sin^2 \theta = 1 - \left(-\frac{1}{4}\right)^2$ $\sin^2 \theta = 1 - \frac{1}{16}$ $\sin^2 \theta = \frac{15}{16}$ Taking the square root, $\sin \theta = \pm \frac{\sqrt{15}}{4}$. Since the angle $\theta$ between two vectors is conventionally taken in the range $[0, \pi]$, $\sin \theta$ is non-negative. $\sin \theta = \frac{\sqrt{15}}{4}$ Reasoning: The magnitude of the cross product formula requires $\sin \theta$. The identity $\sin^2 \theta + \cos^2 \theta = 1$ allows us to find $\sin \theta$ from $\cos \theta$. We take the positive root because the angle between vectors is in $[0, \pi]$.

Step 7: Calculate the Magnitude of the Cross Product $|\vec{a} \times \vec{c}|$. Finally, we use the formula for the magnitude of the cross product: $|\vec{a} \times \vec{c}| = |\vec{a}| |\vec{c}| \sin \theta$. Substitute the known values: $|\vec{a}|=1$, $|\vec{c}|=1$, and $\sin \theta = \frac{\sqrt{15}}{4}$. $|\vec{a} \times \vec{c}| = (1)(1)\left(\frac{\sqrt{15}}{4}\right)$ $|\vec{a} \times \vec{c}| = \frac{\sqrt{15}}{4}$ Reasoning: This is the final step where we combine the magnitudes of the vectors and the sine of the angle between them to find the magnitude of their cross product.

Common Mistakes & Tips

• Squaring Vectors: Remember that $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$, not just $|\vec{v}| \times |\vec{v}|$. When squaring a sum of vectors, use the distributive property of the dot product: $(\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y}) = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$.
• Sign of $\sin \theta$: When calculating the magnitude of a cross product, always use the positive value of $\sin \theta$, as the angle between vectors is in $[0, \pi]$, where $\sin \theta \ge 0$.
• Unit Vector Property: Don't forget to use $|\vec{v}| = 1$ for unit vectors, which simplifies many terms to 1.

Summary The problem was solved by first rearranging the given vector equation to isolate a term involving $\vec{b}$. Squaring both sides allowed us to form an equation involving magnitudes and dot products. By substituting the fact that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors, we solved for the dot product $\vec{a} \cdot \vec{c}$. This dot product was then used to find the cosine of the angle between $\vec{a}$ and $\vec{c}$, and subsequently the sine of the angle using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$. Finally, the magnitude of the cross product $|\vec{a} \times \vec{c}|$ was calculated using its definition.

The final answer is $\boxed{\frac{\sqrt{15}}{4}}$, which corresponds to option (A).