First slide

Scalar triple product of vectors

Question

$\vec{b} and \vec{c}$ are unit vectors. Then for any arbitrary vector $\vec{a}, (((\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})) \times (\vec{b} \times \vec{c}))$ $\cdot (\vec{b} - \vec{c})$ is always equal to

Moderate

A

$| \vec{a} |$
B

$\frac{1}{2} | \vec{a} |$
C

$\frac{1}{3} | \vec{a} |$
D

none of these

Solution

$\begin{array}{l} ((\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})) \times (\vec{b} \times \vec{c}) \\ = (\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c}) + (\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{c}) \\ = ((\vec{a} \times \vec{b}) \cdot \vec{c}) \vec{b} - ((\vec{a} \times \vec{b}) \cdot \vec{b}) \vec{c + ((\vec{a} \times \vec{c}) \cdot \vec{c}) \vec{b} - ((\vec{a} \times \vec{c}) \cdot \vec{b}) \vec{c}} \end{array}$
$\begin{array}{l} = [\vec{a} \vec{b} \vec{c}] (\vec{b} + \vec{c}) \\ \Rightarrow (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})) \times (\vec{b} \times \vec{c})) \cdot (\vec{b} - \vec{c}) \\ = [\vec{a} \vec{b} \vec{c}] (\vec{b} + \vec{c}) \cdot (\vec{b} - \vec{c}) \\ = [\vec{a} \vec{b} \vec{c}] (| \vec{b} |^{2} - | c |^{2}) = 0 \end{array}$

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