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
$\vec{a} + 2\vec{b} - 3\vec{c} = \vec{0} \Rightarrow \vec{a} = -2\vec{b} + 3\vec{c}$
$2\vec{a} \cdot \vec{b} - 6\vec{b} \cdot \vec{c} - 3\vec{c} \cdot \vec{a}$
Substitute $\vec{a} = -2\vec{b} + 3\vec{c}$:
$= 2(-2\vec{b} + 3\vec{c}) \cdot \vec{b} - 6\vec{b} \cdot \vec{c} - 3\vec{c} \cdot (-2\vec{b} + 3\vec{c})$
$= 2(-2\vec{b} \cdot \vec{b} + 3\vec{c} \cdot \vec{b}) - 6\vec{b} \cdot \vec{c} - 3(-2\vec{c} \cdot \vec{b} + 3\vec{c} \cdot \vec{c})$
$= 2(-2 + 3\vec{b} \cdot \vec{c}) - 6\vec{b} \cdot \vec{c} + 6\vec{c} \cdot \vec{b} - 9$
$= -4 + 6\vec{b} \cdot \vec{c} - 6\vec{b} \cdot \vec{c} + 6\vec{b} \cdot \vec{c} - 9$
$= -13 + 6\vec{b} \cdot \vec{c}$
Since $\vec{b} \cdot \vec{c} = \frac{1}{2}(\|\vec{b}\|^2 + \|\vec{c}\|^2 - \|\vec{b} - \vec{c}\|^2)$ and both are unit vectors:
$\vec{b} \cdot \vec{c} \in [-1, 1]$ so maximum value of expression is $-13 + 6(1) = -7$
But from the equation $\vec{a} + 2\vec{b} - 3\vec{c} = 0$, squaring both sides:
$\|\vec{a} + 2\vec{b} - 3\vec{c}\|^2 = 0$
$\vec{a}^2 + 4\vec{b}^2 + 9\vec{c}^2 + 4\vec{a} \cdot \vec{b} - 6\vec{a} \cdot \vec{c} - 12\vec{b} \cdot \vec{c} = 0$
$1 + 4 + 9 + 4\vec{a} \cdot \vec{b} - 6\vec{a} \cdot \vec{c} - 12\vec{b} \cdot \vec{c} = 0$
$14 + 4\vec{a} \cdot \vec{b} - 6\vec{a} \cdot \vec{c} - 12\vec{b} \cdot \vec{c} = 0$
$4\vec{a} \cdot \vec{b} - 6\vec{a} \cdot \vec{c} - 12\vec{b} \cdot \vec{c} = -14$
Divide by 2:
$2\vec{a} \cdot \vec{b} - 3\vec{a} \cdot \vec{c} - 6\vec{b} \cdot \vec{c} = -7$
$\Rightarrow 2\vec{a} \cdot \vec{b} - 6\vec{b} \cdot \vec{c} - 3\vec{c} \cdot \vec{a} = -7$