Consider three vectors a, b, c. Let |a|=2,|b|=3 and a=b c. If [0,… - Vidyadip

MCQ

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

A
110
B
105
C
124
D
121

✓

Answer

$|\overrightarrow{ c }-\overrightarrow{ a }|=|\overrightarrow{ c }|^2+|\overrightarrow{ a }|^2-2 \overline{ a } . \overline{ c }$
$=|\overrightarrow{ c }|^2+4-0$
$\because \overrightarrow{ a }=\overrightarrow{ b } \times \overrightarrow{ c }$
$|\overrightarrow{ a }|=|\overrightarrow{ b } \times \overrightarrow{ c }|$
$2=3|\overrightarrow{ c }| \sin \alpha$
$|\overrightarrow{ c }|=\frac{2}{3} \operatorname{cosec} \alpha \alpha \in\left[0, \frac{\pi}{3}\right]$
$|\overrightarrow{ c }|_{\min }=\frac{2}{3} \times \frac{2}{\sqrt{3}} \operatorname{cosec} \alpha \in\left[\frac{2}{\sqrt{3}}, \infty\right)$
$\Rightarrow 27|\overrightarrow{ c }-\overrightarrow{ a }|_{\min }^2=27\left(\frac{16}{27}+4\right)=124$

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