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If $\vec{a}, \vec{b}$, and $\overrightarrow{\mathrm{c}}$ are unit vectors such that $\vec{a}+2 \vec{b}+2 \overrightarrow{\mathbf{c}}=\overrightarrow{0}$, then $|\vec{a} \times \overrightarrow{\mathbf{c}}|$ is equal to
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$\frac{\sqrt{15}}{4}$
$\frac{\sqrt{15}}{4}$
$\begin{aligned} & \because \vec{a}+2 \vec{b}+2 \overrightarrow{\mathrm{c}}=\overrightarrow{0} \text { [Given] } \\ \Rightarrow & \vec{a}+2 \overrightarrow{\mathrm{c}}=-2 \vec{b} \\ \Rightarrow &(\vec{a}+2 \mathrm{c} \overrightarrow{)} \cdot(\vec{a}+2 \overrightarrow{\mathrm{c}})=(-2 \vec{b})(-2 \overrightarrow{\mathrm{b}})\\ \Rightarrow \vec{a} \cdot \vec{a}+4 \vec{c} \cdot \vec{c}+4 \vec{a} \cdot \vec{c}=4 \vec{b} \cdot \vec{b} \\ \Rightarrow & 1+4+4 \vec{a} \cdot \overrightarrow{\mathrm{c}}=4 \\ \Rightarrow & \vec{a} \cdot \overrightarrow{\mathrm{c}}=\frac{-1}{4} \\ \because &|\vec{a} \cdot \overrightarrow{\mathrm{c}}|^2+|\vec{a} \times \vec{c}|^2=1(\vec{a} \text { is unit vector }) \\ & \frac{1}{16}+|\vec{a} \times \vec{c}|^2=1 \\ &|\vec{a} \times \vec{c}|^2=\frac{15}{16} \\ &|\vec{a} \times \vec{c}|^2=\frac{\sqrt{15}}{4} \end{aligned}$
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