Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a}=\vec{b} imes \vec{c}$. If $\alpha \in\left[0, \frac{\pi}{3} ight]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c}-\vec{a}|^2$ is equal to:

Question

Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a}=\vec{b} 	imes \vec{c}$. If $\alpha \in\left[0, \frac{\pi}{3}ight]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c}-\vec{a}|^2$ is equal to:

Aicharya · Accepted Answer

$\begin{aligned} & |\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{c}}|^2+|\overrightarrow{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}} \ & =|\overrightarrow{\mathrm{c}}|^2+4-0 \ & \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} 	imes \overrightarrow{\mathrm{c}} \ & |\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}} 	imes \overrightarrow{\mathrm{c}}| \ & 2=3|\overrightarrow{\mathrm{c}}| \sin \alpha \ & |\overrightarrow{\mathrm{c}}|=\frac{2}{3} \operatorname{cosec} \alpha \quad \alpha \in\left[0, \frac{\pi}{3}ight] \ & |\overrightarrow{\mathrm{c}}|_{\min }=\frac{2}{3} 	imes \frac{2}{\sqrt{3}} \quad \operatorname{cosec} \alpha \in\left[\frac{2}{\sqrt{3}}, \inftyight) \ & \Rightarrow 27|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|_{\min }^2=27\left(\frac{16}{27}+4ight)=124\end{aligned}$

Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a}=\vec{b}…

Solution