First slide

Vector triple product

Question

If $\vec{a} \times (\vec{b} \times \vec{c})$ is perpendicular to $(\vec{a} \times \vec{b}) \times \vec{c}$ we may have

Difficult

A

$(\vec{a} \cdot \vec{c}) | \vec{b} |^{2} = (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c})$
B

$\vec{a} . \vec{b} = 0$
C

$\vec{a} . \vec{c} = 0$
D

$\vec{b} . \vec{c} = 0$

Solution

$\begin{aligned} \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \\ and (\vec{a} \times \vec{b}) \times \vec{c} = - (\vec{c} \cdot \vec{b}) \vec{a} + (\vec{a} \cdot \vec{c}) \vec{b} \end{aligned}$
We have been given
$\begin{array}{l} (\vec{a} \times (\vec{b} \times \vec{c})) \cdot ((\vec{a} \times \vec{b}) \times \vec{c}) = 0 \\ ∴ ((\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}) \cdot ((\vec{a} \cdot \vec{c}) \vec{b} - (\vec{c} \cdot \vec{b}) \vec{a}) = 0 \\ or (\vec{a} \cdot \vec{c})^{2} | \vec{b} |^{2} - (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{c}) (\vec{a} \cdot \vec{b}) \\ - (\vec{a} \cdot \vec{b}) (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{c}) + (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) (\vec{c} \cdot \vec{a}) = 0 \\ or (\vec{a} \cdot \vec{c})^{2} | \vec{b} |^{2} = (\vec{a} \cdot \vec{c}) (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) \\ or (\vec{a} \cdot \vec{c}) ((\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c})) = 0 \\ \vec{a} \cdot \vec{c} = 0 or (\vec{a} \cdot \vec{c}) | \vec{b} |^{2} = (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) \end{array}$

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