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(B) Given that $\overline{a}, \overline{b}, \overline{c}$ are unit vectors,so $|\overline{a}| = |\overline{b}| = |\overline{c}| = 1$. Given equation:

Vedclass · Accepted Answer

$\frac{\sqrt{15}}{4}$

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(B) Given that $\overline{a}, \overline{b}, \overline{c}$ are unit vectors,so $|\overline{a}| = |\overline{b}| = |\overline{c}| = 1$. Given equation: $\overline{a} + 2\overline{c} = -2\overline{b}$. Squaring both sides: $|\overline{a} + 2\overline{c}|^2 = |-2\overline{b}|^2$ $|\overline{a}|^2 + 4|\overline{c}|^2 + 4(\overline{a} \cdot \overline{c}) = 4|\overline{b}|^2$ Since $|\overline{a}| = |\overline{b}| = |\overline{c}| = 1$,we have: $1 + 4(1) + 4(\overline{a} \cdot \overline{c}) = 4(1)$ $5 + 4(\overline{a} \cdot \overline{c}) = 4$ $4(\overline{a} \cdot \overline{c}) = -1$ $\overline{a} \cdot \overline{c} = -\frac{1}{4}$ Since $\overline{a} \cdot \overline{c} = |\overline{a}||\overline{c}| \cos 	heta = \cos 	heta$,we have $\cos 	heta = -\frac{1}{4}$. Then $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - (-\frac{1}{4})^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$. Finally,$|\overline{a} 	imes \overline{c}| = |\overline{a}||\overline{c}| \sin 	heta = (1)(1) \frac{\sqrt{15}}{4} = \frac{\sqrt{15}}{4}$.

If $\overline{a}, \overline{b}, \overline{c}$ are unit vectors and $\theta$ is the angle between $\overline{a}$ and $\overline{c}$ and $\overline{a}+2 \overline{b}+2 \overline{c}=\overline{0}$,then $|\overline{a} \times \overline{c}|=$

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