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(D) Given that $|\overline{a}| = |\overline{c}| = 1$ and the angle between $\overline{a}$ and $\overline{c}$ is $\frac{\pi}{3}$.$\overline{a} \cdot \o

Vedclass · Accepted Answer

-$50$

Vedclass · Answer

(D) Given that $|\overline{a}| = |\overline{c}| = 1$ and the angle between $\overline{a}$ and $\overline{c}$ is $\frac{\pi}{3}$.$\overline{a} \cdot \overline{c} = |\overline{a}||\overline{c}| \cos \frac{\pi}{3} = 1 	imes 1 	imes \frac{1}{2} = \frac{1}{2}$.Using the vector triple product formula $\overline{a} 	imes (\overline{b} 	imes \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}$.Substituting this into the given equation: $((\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}) \cdot (\overline{a} 	imes \overline{c}) = 5$.Since $(\overline{c} \cdot (\overline{a} 	imes \overline{c})) = 0$ (as the scalar triple product with two identical vectors is zero),the equation simplifies to:$(\overline{a} \cdot \overline{c}) \overline{b} \cdot (\overline{a} 	imes \overline{c}) = 5$.$\frac{1}{2} [\overline{b} \overline{a} \overline{c}] = 5 \Rightarrow [\overline{b} \overline{a} \overline{c}] = 10$.Using the property of scalar triple products,$[\overline{b} \overline{a} \overline{c}] = -[\overline{a} \overline{b} \overline{c}]$.Therefore,$-[\overline{a} \overline{b} \overline{c}] = 10 \Rightarrow [\overline{a} \overline{b} \overline{c}] = -10$.Finally,$5[\overline{a} \overline{b} \overline{c}] = 5 	imes (-10) = -50$.

If $\overline{a}$ and $\overline{c}$ are unit vectors inclined at $\frac{\pi}{3}$ with each other and $(\overline{a} \times (\overline{b} \times \overline{c})) \cdot (\overline{a} \times \overline{c}) = 5$,then the value of $5[\overline{a} \overline{b} \overline{c}] = $

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