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Cross product of vectors or vector product of vectors

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Question

$If \vec{a}, \vec{b} are vectors perpendicular to each other and | \vec{a} | = 2, | \vec{b} | = 3, \vec{c} \times \vec{a} = \vec{b}, then the least value of | \vec{c} - \vec{a} | is$

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Solution

$\begin{array}{l} \vec{c} \times \vec{a} = \vec{b} \\ \Rightarrow | \vec{c} \times \vec{a} | = | \vec{b} | \\ \Rightarrow | \vec{c} | \vec{a} ∣ \sin θ = 3 \\ \Rightarrow | \vec{c} | = \frac{3}{2 \sin θ} \\ \begin{aligned} | \vec{c} - \vec{a} |^{2} & = | \vec{c} |^{2} + | \vec{a} |^{2} - 2 \vec{c} \cdot \vec{a} \\ = | \vec{c} |^{2} + 4 - 2 | \vec{c} | \cdot | \vec{a} | \cos θ \end{aligned} \end{array}$
$\begin{array}{l} = \frac{9}{4 \sin^{2} θ} + 4 - 2 \cdot \frac{3}{2 \sin θ} \cdot 2 \cdot \cos θ \\ = 4 + \frac{9}{4} {cosec}^{2} θ - 6 \cot θ \\ = \frac{9}{4} + {(\frac{3}{2} \cot θ - 2)}^{2} \\ = | \vec{c} - \vec{a} |^{2} \geq \frac{9}{4} \Rightarrow | \vec{c} - \vec{a} | \geq \frac{3}{2} \end{array}$

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