Question

Let three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be such that \(\vec{a}\times\vec{b} = \vec{c}\), \(\vec{b}\times\vec{c} = \vec{a}\) and \(|\vec{a}| = 2\). Then which one of the following is not true?

• A. \(\vec{a}\times\left((\vec{b} + \vec{c})\times(\vec{b}\times\vec{c})\right) = \vec{0}\)
• B. Projection of \(\vec{a}\) on \((\vec{b}\times\vec{c})\) is 2
• C. \(\left[\vec{a}\ \vec{b}\ \vec{c}\right] + \left[\vec{c}\ \vec{a}\ \vec{b}\right] = 8\)
• D. \(\left|3\vec{a} + \vec{b} - 2\vec{c}\right|^2 = 51\)

Hint:

The conditions force \(\vec{a},\vec{b},\vec{c}\) to be mutually perpendicular with \(|\vec{b}| = 1\), \(|\vec{c}| = 2\); then test each statement.

Answer 1

The Correct Option is D

\(\vec{c} = \vec{a}\times\vec{b}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\); \(\vec{a} = \vec{b}\times\vec{c}\) is perpendicular to \(\vec{b}\) and \(\vec{c}\). So \(\vec{a},\vec{b},\vec{c}\) are mutually perpendicular.
\(|\vec{c}| = |\vec{a}||\vec{b}|\) and \(|\vec{a}| = |\vec{b}||\vec{c}| = |\vec{b}|^2|\vec{a}|\), so \(|\vec{b}| = 1\), \(|\vec{c}| = 2\).
(B): \(\vec{b}\times\vec{c} = \vec{a}\), so the projection of \(\vec{a}\) on it is \(|\vec{a}| = 2\), true.
(C): \([\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = |\vec{a}|^2 = 4\) and \([\vec{c}\ \vec{a}\ \vec{b}] = [\vec{a}\ \vec{b}\ \vec{c}] = 4\); sum \(= 8\), true.
(D): since the vectors are perpendicular, \(|3\vec{a} + \vec{b} - 2\vec{c}|^2 = 9(4) + 1 + 4(4) = 53 \ne 51\), so (D) is not true.
Answer: (4) \(\left|3\vec{a} + \vec{b} - 2\vec{c}\right|^2 = 51\)