Question

If $\stackrel{-}{a}+\stackrel{-}{2b}+\stackrel{-}{3c}=\stackrel{-}{0}$, then $\stackrel{-}{a}\times \stackrel{-}{b}+\stackrel{-}{b}\times \stackrel{-}{c}+\stackrel{-}{c}\times \stackrel{-}{a}$ is equal to

Answer 1

$\overline{a}+2\overline{b}+3\overline{c}=0$

$\overline{a}\times (\overline{a}+2\overline{b}+3\overline{c})=0\to 2\overline{a}\times \overline{b}+3\overline{a}\times \overline{c}=0$...(1)

$\overline{b}\times (\overline{a}+2\overline{b}+3\overline{c})=0\to \overline{b}\times \overline{a}+3\overline{b}\times \overline{c}=0\Rightarrow \overline{a}\times \overline{b}=3\overline{b}\times \overline{c}$...(2)

$\overline{c}\times (\overline{a}+2\overline{b}+3\overline{c})=0\to \overline{c}\times \overline{a}+2\overline{c}\times \overline{b}=0\Rightarrow \overline{c}\times \overline{a}=2\overline{b}\times \overline{c}$...(3)

From (2) and (3), we get

$3\overline{c}\times \overline{a}=6\overline{b}\times \overline{c}$

$\Rightarrow \overline{a}\times \overline{b}+\overline{b}\times \overline{c}+\overline{c}\times \overline{a}=3\overline{b}\times \overline{c}+\overline{b}\times \overline{c}+2\overline{b}\times \overline{c}=6\overline{b}\times \overline{c}$

$\therefore \overline{a}\times \overline{b}+\overline{b}\times \overline{c}+\overline{c}\times \overline{a}=6\overline{b}\times \overline{c}$