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#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

**Problem:**

Consider the non zero vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ and $\vec{d}$ such that no three of which are coplanar then prove that $\vec{a}\left[\vec{b} \vec{c} \vec{d}\right]+\vec{c}\left[\vec{a} \vec{b} \vec{d}\right]=\vec{b}\left[\vec{a} \vec{c} \vec{d}\right]+\vec{d}\left[\vec{a} \vec{b} \vec{c}\right]$. Hence prove that if $\vec{a}$, $\vec{b}$, $\vec{c}$ and $\vec{d}$ represent the position vectors of the vertices of a plane quadrilateral then

$$\frac{\left[\vec{a} \vec{b} \vec{d}\right]+\left[\vec{b} \vec{c} \vec{d}\right]}{\left[\vec{a} \vec{b} \vec{c}\right]+\left[\vec{a} \vec{c} \vec{d}\right]}=1$$

**Attempt:**

I am stuck at the first part of the problem. Looking at the two sides, it seems to me that I somehow need to show $\vec{a}\left[\vec{b} \vec{c} \vec{d}\right]$ is same as $\vec{b}\left[\vec{a} \vec{c} \vec{d}\right]$. Since

$$\vec{a}\left[\vec{b} \vec{c} \vec{d}\right]=\vec{a}\left(\vec{b}\cdot \left(\vec{c}\times \vec{d}\right)\right)$$

and

$$\vec{b}\left[\vec{a} \vec{c} \vec{d}\right]=\vec{b} \left(\vec{a}\cdot \left(\vec{c}\times \vec{d}\right)\right)$$

It looks like I need to swap $\vec{a}$ and $\vec{b}$ but that is not a valid step. I don't know how to proceed with the problem.

Any help is appreciated. Thanks!