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Selecting Consecutive Terms in A.P

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Question

If $\vec{a,} \vec{b,} \vec{c}$ are three non-coplanar vectors, then $(\vec{a} + \vec{b} + \vec{c}) . [(\vec{a} + \vec{b}) \times (\vec{a} + \vec{c})]$ equals

(a) 0

(b) $[\vec{a} \vec{b} \vec{c}]$

(c) $2 [\vec{a} \vec{b} \vec{c}]$

(d) $- [\vec{a} \vec{b} \vec{c}]$

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Solution

$(d) - [\vec{a} \vec{b} \vec{c}] We have (\vec{a} + \vec{b} + \vec{c}) . [(\vec{a} + \vec{b}) \times (\vec{a} + \vec{c})] = (\vec{a} + \vec{b} + \vec{c}) . [(\vec{a} + \vec{b}) \times \vec{a} + (\vec{a} + \vec{b}) \times \vec{c}] (By definition of cross poduct) = (\vec{a} + \vec{b} + \vec{c}) . [\vec{a} \times \vec{a} + \vec{b} \times \vec{a} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c}] = (\vec{a} + \vec{b} + \vec{c}) . [0 + \vec{b} \times \vec{a} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c}] = \vec{a} . (\vec{b} \times \vec{a}) + \vec{a} . (\vec{a} \times \vec{c}) + \vec{a} . (\vec{b} \times \vec{c}) + \vec{b} . (\vec{b} \times \vec{a}) + \vec{b} . (\vec{a} \times \vec{c}) + \vec{b} . (\vec{b} \times \vec{c}) + \vec{c} . (\vec{b} \times \vec{a}) + \vec{c} . (\vec{a} \times \vec{c}) + \vec{c} . (\vec{b} \times \vec{c}) = [\vec{a} \vec{b} \vec{a}] + [\vec{a} \vec{a} \vec{c}] + [\vec{a} \vec{b} \vec{c}] + [\vec{b} \vec{b} \vec{a}] + [\vec{b} \vec{a} \vec{c}] + [\vec{b} \vec{b} \vec{c}] + [\vec{c} \vec{b} \vec{a}] + [\vec{c} \vec{a} \vec{c}] + [\vec{c} \vec{b} \vec{c}] = 0 + 0 + [\vec{a} \vec{b} \vec{c}] + 0 + [\vec{b} \vec{a} \vec{c}] + 0 + [\vec{c} \vec{b} \vec{a}] + 0 + 0 = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] (∵ [\vec{b} \vec{a} \vec{c}] = - [\vec{c} \vec{a} \vec{b}], [\vec{b} \vec{a} \vec{c}] = - [\vec{a} \vec{b} \vec{c}]) = - [\vec{a} \vec{b} \vec{c}]$

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Similar questions

\to a, \to b

\to c

are three non – coplanar vectors, then

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= _______________.

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