Consider three vectors overrightarrow{a}, ove | vedclass

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Question

(C) Given $\overrightarrow{a} = \overrightarrow{b} 	imes \overrightarrow{c}$,it implies $\overrightarrow{a} \perp \overrightarrow{b}$ and $\overright

Vedclass · Accepted Answer

$124$

Vedclass · Answer

(C) Given $\overrightarrow{a} = \overrightarrow{b} 	imes \overrightarrow{c}$,it implies $\overrightarrow{a} \perp \overrightarrow{b}$ and $\overrightarrow{a} \perp \overrightarrow{c}$. Thus,$\overrightarrow{a} \cdot \overrightarrow{c} = 0$.We have $|\overrightarrow{c} - \overrightarrow{a}|^2 = |\overrightarrow{c}|^2 + |\overrightarrow{a}|^2 - 2(\overrightarrow{a} \cdot \overrightarrow{c}) = |\overrightarrow{c}|^2 + 2^2 - 0 = |\overrightarrow{c}|^2 + 4$.From $|\overrightarrow{a}| = |\overrightarrow{b} 	imes \overrightarrow{c}|$,we get $2 = |\overrightarrow{b}| |\overrightarrow{c}| \sin \alpha = 3 |\overrightarrow{c}| \sin \alpha$.Therefore,$|\overrightarrow{c}| = \frac{2}{3 \sin \alpha} = \frac{2}{3} \csc \alpha$.Since $\alpha \in [0, \frac{\pi}{3}]$,the range of $\csc \alpha$ is $[\frac{2}{\sqrt{3}}, \infty)$.The minimum value of $|\overrightarrow{c}|$ occurs at $\alpha = \frac{\pi}{3}$,where $|\overrightarrow{c}| = \frac{2}{3} 	imes \frac{2}{\sqrt{3}} = \frac{4}{3\sqrt{3}}$.Thus,$|\overrightarrow{c}|^2 = \frac{16}{27}$.Substituting this into the expression,$27|\overrightarrow{c} - \overrightarrow{a}|^2 = 27(|\overrightarrow{c}|^2 + 4) = 27(\frac{16}{27} + 4) = 16 + 108 = 124$.

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