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Scalar triple product of vectors

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Question

Let $\vec{u} and \vec{v}$ be unit vectors such that $\vec{u} \times \vec{v} + \vec{u} = \vec{w} and \vec{w} \times \vec{u} = \vec{v}$ . Find the number of $[\vec{u} \vec{v} \vec{w}]$

Moderate

Solution

$Given, \vec{u} \times \vec{v} + \vec{u} = \vec{w} and \vec{w} \times \vec{u} = \vec{v}$
$\begin{array}{l} \Rightarrow \vec{(u} \times \vec{v} + \vec{u}) \times \vec{u} = \vec{v} \\ \Rightarrow (\vec{u} \times \vec{v}) \times \vec{u} = \vec{v} \\ \Rightarrow \vec{v} - (\vec{u} \cdot \vec{v}) = \vec{v} \\ \Rightarrow (\vec{u} \cdot \vec{v}) \vec{u} = 0 \Rightarrow (\vec{u} \vec{v}) = 0 \end{array}$
$Now, [\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$
$\begin{array}{l} = \vec{u} \cdot (\vec{v} \times (\vec{u} \times \vec{v} + \vec{u})) \\ = \vec{u} \cdot (\vec{v} \times (\vec{u} \times \vec{v}) + \vec{v} \times \vec{u}) \\ = \vec{u} ({\vec{v}}^{2} \vec{u} - (\vec{u} \cdot \vec{v}) \vec{v} + \vec{v} \times \vec{u}) = \vec{v^{2}} \vec{u^{2}} = 1 \end{array}$

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Similar Questions

$[(\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c}) (\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b})]$ is equal to (where $\vec{a}, \vec{b} and \vec{c}$ are non-zero non-coplanar vectors)

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