Consider three vectors a,b,c.

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 $\alpha \in [0,\frac{\pi }{3}]$ 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:

Accepted Answer

$\begin{gathered} \vec{a}=\vec{b}	imes \vec{c} \ |\vec{a}|=2,|\vec{b}|=3 \end{gathered}$

$\vec{a}\cdot \vec{b}=0$ and $\vec{a}\cdot \vec{c}=0$

$\begin{gathered} |\vec{c}-\vec{a}{|}^{2}=|\vec{c}{|}^{2}+|\vec{a}{|}^{2}-2\vec{c}\cdot \vec{a} \ =4+|\vec{c}{|}^{2} \end{gathered}$

$\begin{gathered} |\vec{a}|=|\vec{b}	imes \vec{c}|=|\vec{b}|\sin \alpha |\vec{c}| \ \Rightarrow \sin \alpha |\vec{c}|=\frac{2}{3} \ \Rightarrow {\sin }^{2}\alpha =\frac{4}{9|\vec{c}{|}^{2}} \ \Rightarrow |\vec{c}{|}^{2}=\frac{4}{9{\sin }^{2}\alpha } \ \Rightarrow |\vec{c}-\vec{a}{|}^{2}=4+\frac{4}{9{\sin }^{2}\alpha } \end{gathered}$

For $|\vec{c}-\vec{a}{|}^{2}$ to be minimum for $\alpha \in [0,\frac{\pi }{3}]$

$\begin{gathered} \sin \alpha =\frac{\sqrt{3}}{2} \ 27|\vec{c}-\vec{a}{|}^{2}=27[4+\frac{4.4}{9.3}] \ =27[\frac{124}{27}]=124 \end{gathered}$

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