Consider three vectors . Let and . If is the angle between the vectors and , then the minimum value of is equal to : [2024]

Question

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 $α \in [0, \frac{π}{3}]$ 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 : [2024]

Smart Achievers · Accepted Answer

(2)

Consider $| \vec{c} - \vec{a} |^{2}$

$= | \vec{c} |^{2} + | \vec{a} |^{2} - 2 \vec{a} \cdot \vec{c} = | \vec{c} |^{2} + 4 - 2 (\vec{b} \times \vec{c}) \cdot \vec{c} = | \vec{c} |^{2} + 4 - 0$

$∴ | \vec{c} - \vec{a} |^{2} = | \vec{c} |^{2} + 4$ ... (i)

Now, $| \vec{a} | = | \vec{b} \times \vec{c} | \Rightarrow 2 = | \vec{b} | | \vec{c} | \sin α, α \in [0, \frac{π}{3}]$

$= 3 | \vec{c} | \sin α$

$\Rightarrow | \vec{c} | = \frac{2}{3} cosec α$

$\Rightarrow | \vec{c} |_{\min} = \frac{2}{3} \times \frac{2}{\sqrt{3}}$ $(∵ α \in [0, \frac{π}{3}])$

$\Rightarrow 27 | \vec{c} - \vec{a} |_{\min}^{2} = 27 (\frac{16}{27} + 4)$

=16 + 108 = 124.

Solution

1 105

2 124

3 121

4 110

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