Let \( \vec{a}, \vec{b}, \vec{c} \) be unit vectors such that \( \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \). Which one of the following is correct ? [IIT-2007] (a) \( \vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}=\overrightarrow{0} \) (b) \( \overrightarrow{ a } \times \overrightarrow{ b }=\overrightarrow{ b } \times \overrightarrow{ c }=\overrightarrow{ c } \times \overrightarrow{ a } \neq \overrightarrow{0} \) (c) \( \overrightarrow{ a } \times \overrightarrow{ b }=\overrightarrow{ b } \times \overrightarrow{ c }=\overrightarrow{ a } \times \overrightarrow{ c } \neq \overrightarrow{0} \) (d) \( \vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a} \) are mutually \( \perp \)

Given that

\(\vec a, \vec b\) and \(\vec c\)are unit vectors.

i.e., |\(\vec a\)| = |\(\vec b\)|  = |\(\vec c\)| = 1

And \(\vec a\) + \(\vec b\) + \(\vec c\) = \(\vec 0\)

Then \((\vec a+\vec b+\vec c).(\vec a+\vec b+\vec c)=\vec 0.\vec 0=0\)

⇒ \(\vec a\).\(\vec a\) + \(\vec b\).\(\vec b\) +\(\vec c\).\(\vec c\) + 2(\(\vec a\).\(\vec b\) +\(\vec b\).\(\vec c\) + \(\vec c\). \(\vec a\)) = 0

⇒ \(\vec a\).\(\vec b\) + \(\vec b\).\(\vec c\) + \(\vec c\).\(\vec a\) = -3/2 (\(\because \vec a.\vec a=|\vec a|^2=1 \)))

\(\therefore\vec a,\vec b\)and \(\vec c\) are not parallel to each other.

Now,

\(\vec a\times\vec b=\vec a\times(-(\vec a+\vec c))\) \((\because\vec b=-(\vec a+\vec c))\)

\(=-\vec a\times\vec a-\vec a\times\vec c\) 

\(=-\vec a\times\vec c\) \((\because\vec a\times\vec a=\vec v)\)

\(=\vec c\times\vec a\)

And \(\vec b\times\vec c=-(\vec a+\vec c)\times\vec c\)

\(=-\vec a\times\vec c-\vec c\times\vec c\)

\(=\vec c\times\vec a(\because\vec c\times\vec c=\vec 0)\)

\(\therefore\vec a\times\vec b=\vec c\times\vec a=\vec b\times\vec c\neq\vec 0\)

(Because \(\vec a\),\(\vec b\) and \(\vec c\) are not mutually parallel)

Option (b) is correct