Question

If $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}$, then $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ is equal to ?

• A. $abc$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$

$Given,\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}weknowthat-\overrightarrow{a}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{c}=\overrightarrow{0}⟶\left(1\right)\overrightarrow{a}\times \overrightarrow{b}=-(\overrightarrow{b}\times \overrightarrow{a}),\overrightarrow{b}\times \overrightarrow{c}=-(\overrightarrow{c}\times \overrightarrow{b}),\overrightarrow{c}\times \overrightarrow{a}=-(\overrightarrow{a}\times \overrightarrow{c})⟶\left(2\right)Takingfirsttwo-\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}\Rightarrow \overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{0}\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{b}=\overrightarrow{0}using\left(2\right)\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{b}=\overrightarrow{0}using\left(1\right)\because (\overrightarrow{b}\times \overrightarrow{b}=\overrightarrow{0})\Rightarrow (\overrightarrow{a}+\overrightarrow{c}+\overrightarrow{b})\times \overrightarrow{b}=\overrightarrow{0}usingdistributivelaw.\Rightarrow (\overrightarrow{a}+\overrightarrow{c}+\overrightarrow{b})=\overrightarrow{0}as\overrightarrow{b}\ne \overrightarrow{0}\Rightarrow (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{0}⟶\left(3\right)Similarly,takingsecondandthirdterms\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}\Rightarrow \overrightarrow{b}\times \overrightarrow{c}-\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0}\Rightarrow \overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{0}using\left(2\right)\Rightarrow \overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{c}=\overrightarrow{0}using\left(1\right)\because (\overrightarrow{c}\times \overrightarrow{c}=\overrightarrow{0})\Rightarrow (\overrightarrow{b}+\overrightarrow{a}+\overrightarrow{c})\times \overrightarrow{c}=\overrightarrow{0}usingdistributivelaw\Rightarrow (\overrightarrow{b}+\overrightarrow{a}+\overrightarrow{c})=\overrightarrow{0}as\overrightarrow{c}\ne \overrightarrow{0}\Rightarrow (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{0}⟶\left(4\right)Similarly,takingfirstandthirdterms\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a}\Rightarrow \overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0}\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{0}using\left(2\right)\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{a}\times \overrightarrow{a}=\overrightarrow{0}using\left(1\right)\because (\overrightarrow{a}\times \overrightarrow{a}=\overrightarrow{0})\Rightarrow \overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{a})=\overrightarrow{0}usingdistributivelaw.\Rightarrow (\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{a})=\overrightarrow{0}as\overrightarrow{a}\ne \overrightarrow{0}\Rightarrow (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{0}⟶\left(5\right)\therefore from\left(3\right),\left(4\right)and\left(5\right)weget(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{0}$