For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) prove that [\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] = 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]

Solution :

We have [\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)]

= {(\(\vec{a}\) + \(\vec{b}\))\(\times\)(\(\vec{b}\) + \(\vec{c}\))}.(\(\vec{c}\) + \(\vec{a}\))

= {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{b}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\))  {\(\vec{b}\)\(\times\)\(\vec{b}\) = 0}

= {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\))

= (\(\vec{a}\times\vec{b}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{b}\)).\(\vec{a}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{a}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{a}\)

= [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] + 0 + 0 + 0 + 0 + [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)]

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


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