Three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) satisfy the condition \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\). Evaluate the quantity \(\mu = \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}\), if \(|\vec{a}|=3\), \(|\vec{b}|=4\) and \(|\vec{c}|=2\).

$-\frac{29}{2}$

The correct answer is Option (3) → $-\frac{29}{2}$ ##

Since \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), we have

$\vec{a} + \vec{b} + \vec{c} = \vec{0}$

or $\vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = 0$

Therefore $\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = -|\vec{a}|^2 = -9 \quad \dots (1)$

Again, $\vec{b}\cdot(\vec{a} + \vec{b} + \vec{c}) = 0$

or $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} = -|\vec{b}|^2 = -16 \quad \dots (2)$

Similarly, $\vec{c}\cdot\vec{a} + \vec{b}\cdot\vec{c} = -4 \quad \dots (3)$

Adding (1), (2) and (3), we have

$2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = -9 -16 -4 = -29$

or $2\mu = -29 \quad \Rightarrow \quad \mu = -\frac{29}{2}$