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If \(\vec{a}+\vec{b}+\vec{c}=0,\) then prove that \(\vec{a}\times \vec{b}=\vec{b}\times \vec{c}=\vec{c}\times \vec{a}\)

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[By the distributive law]

[By the distributive law]

From (i) and (ii), we get \((\vec{a}\times \vec{b})=(\vec{b}\times \vec{c})=(\vec{c}\times \vec{a}).\)

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