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Step-by-Step Solution
Step 1: Express the Given Condition
We are given that
$ \overrightarrow{a} + 2\overrightarrow{b} + 2\overrightarrow{c} = \overrightarrow{0}. $
From this, we can rearrange as
$ \overrightarrow{a} + 2\overrightarrow{c} = -2\overrightarrow{b}. $
Step 2: Square Both Sides
Taking the magnitude on both sides and then squaring, we get:
$ |\overrightarrow{a} + 2\overrightarrow{c}|^2 = |-2\overrightarrow{b}|^2. $
Step 3: Expand the Magnitude Squared Terms
Recall that
$ |\mathbf{x} + \mathbf{y}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2 + 2\,\mathbf{x}\cdot\mathbf{y}. $
So,
$ |\overrightarrow{a}|^2 + 4|\overrightarrow{c}|^2 + 4\,\overrightarrow{a} \cdot \overrightarrow{c} = 4|\overrightarrow{b}|^2. $
Step 4: Substitute the Unit Vector Magnitudes
Since $ \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} $ are unit vectors, we know
$ |\overrightarrow{a}|^2 = |\overrightarrow{b}|^2 = |\overrightarrow{c}|^2 = 1. $
Hence,
$ 1 + 4 \times 1 + 4(\overrightarrow{a} \cdot \overrightarrow{c}) = 4 \times 1. $
Thus,
$ 1 + 4 + 4(\overrightarrow{a} \cdot \overrightarrow{c}) = 4. $
Step 5: Solve for the Dot Product
Simplifying,
$ 5 + 4(\overrightarrow{a} \cdot \overrightarrow{c}) = 4, $
which gives
$ 4(\overrightarrow{a} \cdot \overrightarrow{c}) = -1, $
so
$ \overrightarrow{a} \cdot \overrightarrow{c} = -\frac{1}{4}. $
Step 6: Relate the Dot Product to the Angle
For vectors $ \overrightarrow{a} $ and $ \overrightarrow{c} $ of unit magnitude,
$ \overrightarrow{a} \cdot \overrightarrow{c} = |\overrightarrow{a}||\overrightarrow{c}|\cos\theta = \cos\theta. $
Thus,
$ \cos\theta = -\frac{1}{4}. $
Step 7: Find $ \sin\theta $ Using the Identity
Using $ \sin^2\theta = 1 - \cos^2\theta, $
we get
$ \sin^2\theta = 1 - \left(-\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16}. $
Therefore,
$ \sin\theta = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}. $
Step 8: Find the Magnitude of the Cross Product
The magnitude of the cross product is
$ |\overrightarrow{a} \times \overrightarrow{c}| = |\overrightarrow{a}||\overrightarrow{c}|\sin\theta. $
Since $|\overrightarrow{a}| = 1$ and $|\overrightarrow{c}| = 1$, we have
$ |\overrightarrow{a} \times \overrightarrow{c}| = 1 \cdot 1 \cdot \frac{\sqrt{15}}{4} = \frac{\sqrt{15}}{4}. $
Final Answer
$ \displaystyle \frac{\sqrt{15}}{4}. $
