Question

If the vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ from the sides $BC,CA$ and $AB$ respectively, of a triangle $ABC$, then

• A. $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=0$
• B. $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}$
• C. $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$
• D. $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=0$

Answer 1

Answer: (B)

By triangle law, $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$

Taking cross product by $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ respectively,
$\overrightarrow{a}\times (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times \overrightarrow{0}=\overrightarrow{0}$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{a}$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a}[\because \overrightarrow{a}\times \overrightarrow{a}=\overrightarrow{o}]$
Similarly, $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{a}$
$\therefore \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}\times \overrightarrow{a}$