Question

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be unit vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}.$ Which one of the following is correct?

• A. $\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{c}\times\overrightarrow{a}=\overrightarrow{0}$
• B. $\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{c}\times\overrightarrow{a}\ne \overrightarrow{0}$
• C. $\overrightarrow{b}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{0}$
• D. $\overrightarrow{a}\times\overrightarrow{b},\overrightarrow{b}\times\overrightarrow{c},\overrightarrow{c}\times\overrightarrow{a}$ are mutually perpendicular

Answer 1

Answer: (B)

$\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{c}\times\overrightarrow{a}\ne \overrightarrow{0}$

Answer 2

Since, $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors and $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}=\overrightarrow{0}$
$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ represent an equilateral triangle.
\therefore \hspace25mm \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{c}\times\overrightarrow{a}\ne \overrightarrow{0}.