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Vector algebra problem #2

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!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,715
Hint: Use Lagrange's formula to expand $(\vec a \times\vec b)\times (\vec c\times \vec d) = -(\vec c\times \vec d)\times (\vec a \times\vec b).$
 

Pranav

Well-known member
Nov 4, 2013
428
Hint: Use Lagrange's formula to expand $(\vec a \times\vec b)\times (\vec c\times \vec d) = -(\vec c\times \vec d)\times (\vec a \times\vec b).$
Hi Opalg! :)

Let $(\vec a \times\vec b)=\vec{u}$. Then from the Lagrange's formula, I have:
$$\vec{u}\times (\vec{c}\times \vec{d})=\vec{c}(\vec{u}\cdot \vec{d})-\vec{d}(\vec{u}\cdot \vec{c})=\vec{c}\left[\vec{a} \vec{b} \vec{d}\right]-\vec{d}\left[\vec{a} \vec{b} \vec{c}\right]$$
Similarly, I have:
$$(\vec c\times \vec d)\times (\vec a \times\vec b)=\vec{a}\left[\vec{b} \vec{c} \vec{d}\right]-\vec{b}\left[\vec{a} \vec{c} \vec{d}\right]$$
Since $(\vec a \times\vec b)\times (\vec c\times \vec d) = -(\vec c\times \vec d)\times (\vec a \times\vec b)$, I get:
$$\vec{c}\left[\vec{a} \vec{b} \vec{d}\right]-\vec{d}\left[\vec{a} \vec{b} \vec{c}\right]=-\vec{a}\left[\vec{b} \vec{c} \vec{d}\right]+\vec{b}\left[\vec{a} \vec{c} \vec{d}\right]$$
$$\Rightarrow \vec{a}\left[\vec{b} \vec{c} \vec{d}\right]+\vec{c}\left[\vec{a} \vec{b} \vec{d}\right]=\vec{b}\left[\vec{a} \vec{c} \vec{d}\right]+\vec{d}\left[\vec{a} \vec{b} \vec{c}\right]$$
which proves the first part.

How to proceed with the second one? :confused:
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,715
How to proceed with the second one? :confused:
I was hoping you wouldn't ask that, because I am not sure of the answer. (Thinking)

The condition for a quadrilateral to be planar is that its diagonals should intersect. In terms of the given vectors, this means that there should exist scalars $\lambda,\mu$ such that $\lambda\vec a + (1-\lambda)\vec c = \mu\vec b + (1-\mu)\vec d$. The first part of the problem says that there is some linear combination of $\vec a$ and $\vec c$ that is equal to some linear combination of $\vec b$ and $\vec d$. There must be some way of relating those two pieces of information. (Wondering)
 

Pranav

Well-known member
Nov 4, 2013
428
I was hoping you wouldn't ask that, because I am not sure of the answer. (Thinking)

The condition for a quadrilateral to be planar is that its diagonals should intersect. In terms of the given vectors, this means that there should exist scalars $\lambda,\mu$ such that $\lambda\vec a + (1-\lambda)\vec c = \mu\vec b + (1-\mu)\vec d$. The first part of the problem says that there is some linear combination of $\vec a$ and $\vec c$ that is equal to some linear combination of $\vec b$ and $\vec d$. There must be some way of relating those two pieces of information. (Wondering)
Ah but many thanks for the help you provided so far. :)

I figured out the second part but I couldn't use any of the result from the first part. (Thinking)

Here is what I did:

Since the vectors form a plane quadrilateral, the vectors $\vec{b}-\vec{a}$, $\vec{c}-\vec{a}$ and $\vec{d}-\vec{a}$ must be coplanar, hence I have the condition:

$$\left[\vec{b}-\vec{a}\,\,\, \vec{c}-\vec{a}\,\,\, \vec{d}-\vec{a}\right]=0$$
$$\Rightarrow (\vec{b}-\vec{a})\cdot ((\vec{c}-\vec{a})\times (\vec{d}-\vec{a}))=0$$
$$\Rightarrow (\vec{b}-\vec{a})\cdot (\vec{c}\times \vec{d} +\vec{a}\times \vec{c}+\vec{d} \times \vec{a})=0$$
$$\Rightarrow \left[\vec{b} \vec{c} \vec{d}\right]+\left[\vec{b} \vec{a} \vec{c}\right]+\left[\vec{b} \vec{d} \vec{a}\right]-\left[\vec{a} \vec{c} \vec{d}\right]=0$$
I can write $\left[\vec{b} \vec{d} \vec{a}\right]$ as $\left[\vec{a} \vec{b} \vec{d}\right]$ and $\left[\vec{b} \vec{a} \vec{c}\right]$ as $-\left[\vec{a} \vec{b} \vec{c}\right]$. From here, second part can proved now.

Thanks again! :)