If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are three vectors such that $|\mathbf{a}|=|\mathbf{b}|=2,…
If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are three vectors such that $|\mathbf{a}|=|\mathbf{b}|=2, \mathbf{a} \cdot \mathbf{b}=2$ and $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$, then $|\mathbf{c}|$ is equal to
2
$2 \sqrt{3}$
$\sqrt{3}$
3
Solution
Given, |a |= |b|= 2, a ×b = 2
and a + b + c = 0, then|c|= ?
a + b + c = 0 … (i)
$\begin{aligned} & \Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=0 \\ & \Rightarrow|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})=0 \\ & 2^2+2^2+|\mathbf{c}|^2+2[2+\mathbf{c}(\mathbf{a}+\mathbf{b})]=0 \\ & \Rightarrow 4+4+|\mathbf{c}|^2+2[2+\mathbf{c}(-\mathbf{c})]=0 \\ & \text { \{from Eq. (i), } \mathbf{a}+\mathbf{b}=-\mathbf{c}\} \\ & \Rightarrow \quad 8+|\mathbf{c}|^2+4-2|\mathbf{c}|^2=0 \\ & \Rightarrow \quad|\mathbf{c}|^2=12 \Rightarrow|\mathbf{c}|=2 \sqrt{3} \\ & \end{aligned}$