Question

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be three vectors in the $xyz$ space such that $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}\ne \overrightarrow{0}$.
If $A,B,C$ are points with position vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ respectively, then the number of possible positions of the centroid of triangle $ABC$ is.

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is A $1$

$\left(\underset{a}{\to \times }\underset{b}{\to }\right)=(\underset{b}{\to }\times \underset{c}{\to })=(\underset{c}{\to }\times \underset{a}{\to }).......\left(i\right)\Rightarrow (\underset{a}{\to }\times \underset{b}{\to })-(\underset{b}{\to }\times \underset{c}{\to })=0(\underset{a}{\to }\times \underset{b}{\to })+(\underset{c}{\to }\times \underset{b}{\to })=0(\underset{a}{\to }+\underset{c}{\to })\times \underset{b}{\to }=0\Rightarrow (\underset{a}{\to }+\underset{c}{\to })=\lambda \left(\underset{b}{\to }\right)\Rightarrow \underset{b}{\to }=\frac{(\underset{a}{\to }+\underset{c}{\to })}{\lambda }......\left(ii\right)$

$(\underset{b}{\to }\times \underset{c}{\to })=(\underset{c}{\to }\times \underset{a}{\to })$

using$\left(ii\right)$

$(\frac{(\underset{a}{\to }+\underset{c}{\to })}{\lambda }\times \underset{c}{\to })=(\underset{c}{\to }\times \underset{a}{\to })\Rightarrow (\underset{a}{\to }\times \underset{c}{\to })=\lambda (\underset{c}{\to }\times \underset{a}{\to })(\underset{a}{\to }\times \underset{c}{\to })+\lambda (\underset{a}{\to }\times \underset{c}{\to })=0(1+\lambda )(\underset{a}{\to }\times \underset{c}{\to })=0\Rightarrow \lambda =-1$

From $\left(i\right)$

$(\underset{a}{\to }+\underset{c}{\to })=-\underset{b}{\to }\Rightarrow (\underset{a}{\to }+\underset{b}{\to }+\underset{c}{\to })=0$

Centriod of taiangle = $\frac{(\underset{a}{\to }+\underset{b}{\to }+\underset{c}{\to })}{3}=\frac{0}{3}=0$

So option $A$ is correct