First slide

Dot product or scalar product of vectors

Question

$If \vec{a} and \vec{b} are perpendicular unit vectors and vector \vec{c} is such that \vec{c} = \vec{a} + \vec{b}, then the value of$
$(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) is$

Moderate

Solution

$\begin{array}{l} (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) \\ = (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{b}) + (\vec{b} \cdot \vec{c}) (\vec{c} \cdot \vec{a}) - (\vec{b} \cdot \vec{a}) (\vec{c} \cdot \vec{c}) + (\vec{c} \cdot \vec{a}) (\vec{a} \cdot \vec{b}) - (\vec{c} \cdot \vec{b}) (\vec{a} \cdot \vec{a}) \end{array}$
$= 0 - (\vec{a} \cdot \vec{c}) + (\vec{b} \cdot \vec{c}) (\vec{c} \cdot \vec{a}) - 0 + 0 - (\vec{c} \cdot \vec{b}) (∵ \vec{a} \cdot \vec{b} = 0)$
$\begin{array}{l} = (\vec{b} \cdot \vec{c}) (\vec{c} \cdot \vec{a}) - (\vec{c} \cdot \vec{a}) - (\vec{b} \cdot \vec{c}) + 1 - 1 \\ = ((\vec{b} \cdot \vec{c}) - 1) ((\vec{c} \cdot \vec{a}) - 1) - 1 \\ = (1 - 1) (1 - 1) - 1 = - 1 (∵ \vec{c} = \vec{a} + \vec{b}) \end{array}$

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