If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three non zero vectors such that \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\), prove that \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually at right angles and |\(\vec{b}\)| = 1 and |\(\vec{c}\)| = |\(\vec{a}\)|

Solution :

\(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\)

\(\implies\)  \(\vec{c}\perp\vec{a}\) , \(\vec{c}\perp\vec{b}\) and \(\vec{a}\perp\vec{b}\), \(\vec{a}\perp\vec{c}\)

\(\implies\)  \(\vec{a}\perp\vec{b}\), \(\vec{b}\perp\vec{c}\) and \(\vec{c}\perp\vec{a}\)

\(\implies\)  \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors.

Again, \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\)

\(\implies\) |\(\vec{a}\times\vec{b}\)| = |\(\vec{c}\)| and |\(\vec{b}\times\vec{c}\)| = |\(\vec{a}\)|

\(\implies\)  \(|\vec{a}||\vec{b}|sin{\pi\over 2}\) = |\(\vec{c}\)| and \(|\vec{b}||\vec{c}|sin{\pi\over 2}\) = |\(\vec{a}\)|  (\(\because\) \(\vec{a}\perp\vec{b}\) and \(\vec{b}\perp\vec{c}\))

\(\implies\)  \(|\vec{a}||\vec{b}|\) = |\(\vec{c}\)| and \(|\vec{b}||\vec{c}|\) = |\(\vec{a}\)|

\(\implies\)  \({|\vec{b}|}^2\) |\(\vec{c}\)| = |\(\vec{c}\)|

\(\implies\)  \({|\vec{b}|}^2\) = 1

\(\implies\)  \(|\vec{b}|\) = 1

putting in \(|\vec{a}||\vec{b}|\) = |\(\vec{c}\)|

\(\implies\)  \(|\vec{a}|\) = |\(\vec{c}\)|


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