# Prove the following

Question:

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such

that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$. if

$\lambda=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ and

$\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}$, then

the ordered pair, $(\lambda, \vec{d})$ is equal to:

1. $\left(\frac{3}{2}, 3 \vec{a} \times \vec{c}\right)$

2. $\left(-\frac{3}{2}, 3 \vec{c} \times \vec{b}\right)$

3. $\left(\frac{3}{2}, 3 \vec{b} \times \vec{c}\right)$

4. $\left(-\frac{3}{2}, 3 \vec{a} \times \vec{b}\right)$

Correct Option: , 4

Solution:

$|\vec{a}+\vec{b}+\vec{c}|^{2}=0$

$3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$

$(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=\frac{-3}{2} \Rightarrow \lambda=\frac{-3}{2}$

$\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times(-\vec{a}-\vec{b})+(-\vec{a}-\vec{b}) \times \vec{a}[\because \vec{c}=-\vec{a}-\vec{b}]$

$=\vec{a} \times \vec{b}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}$

$\vec{d}=3(\vec{a} \times \vec{b})$