First slide

Vector triple product

Question

If $\vec{a} \times \vec{b} = \vec{c}, \vec{b} \times \vec{c} = \vec{a}, where \vec{c} \neq \vec{0}$ then

Moderate

A

$| \vec{a} | = | \vec{c} |$
B

$| \vec{a} | = | \vec{b} |$
C

$| \vec{b} | = 1$
D

$| \vec{a} | = | \vec{b} | = | \vec{c} | = 1$

Solution

$\vec{a} \times \vec{b} = \vec{c}, \vec{b} \times \vec{c} = \vec{a}$
Taking cross with $\vec{b}$ in the first equation, we get
$\begin{array}{ll} \vec{b} \times (\vec{a} \times \vec{b}) = \vec{b} \times \vec{c} = \vec{a} \\ or | \vec{b} |^{2} \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} = \vec{a} \\ \Rightarrow | \vec{b} | = 1 and \vec{a} \cdot \vec{b} = 0 \\ Also | \vec{a} \times \vec{b} | = | \vec{c} | or | \vec{a} | | \vec{b} | \sin \frac{π}{2} = | \vec{c} | \\ or | \vec{a} | = | \vec{c} | \end{array}$

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