If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three unit vectors such that \(\vec{a} + 2\vec{b} - 3\vec{c} = \vec{0}\), then the value of \(2\vec{a}\cdot\vec{b} - 6\vec{b}\cdot\vec{c} - 3\vec{c}\cdot\vec{a}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-7\)
Each dot product comes from keeping two of the vectors on one side of the given relation and taking the square of the magnitude of both sides, with \(|\vec{a}| = |\vec{b}| = |\vec{c}| = 1\).
From \(\vec{a} + 2\vec{b} = 3\vec{c}\):
\(1 + 4 + 4\,\vec{a}\cdot\vec{b} = 9\)
\(\vec{a}\cdot\vec{b} = 1\)
From \(\vec{a} - 3\vec{c} = -2\vec{b}\):
\(1 + 9 - 6\,\vec{c}\cdot\vec{a} = 4\)
\(\vec{c}\cdot\vec{a} = 1\)
From \(2\vec{b} - 3\vec{c} = -\vec{a}\):
\(4 + 9 - 12\,\vec{b}\cdot\vec{c} = 1\)
\(\vec{b}\cdot\vec{c} = 1\)
So
\(2\vec{a}\cdot\vec{b} - 6\vec{b}\cdot\vec{c} - 3\vec{c}\cdot\vec{a} = 2 - 6 - 3 = -7\)
(A dot product of 1 between unit vectors means they point the same way, so here \(\vec{a} = \vec{b} = \vec{c}\), and indeed \(1 + 2 - 3 = 0\).)