# vector problem - exam revision

• Jan 6th 2010, 06:17 AM
schteve
vector problem - exam revision
I have 3 vectors a b and c such that a+b+c=0. This implies they are all the same length as they finish where they start. The question then asks me to prove that axb = bxc = cxa. This seems obvious as if the values are the same then the cross products will be too. However i need to prove this and am not sure how to. Any help would be appreciated.
• Jan 6th 2010, 07:29 AM
earboth
Quote:

Originally Posted by schteve
I have 3 vectors a b and c such that a+b+c=0. This implies they are all the same length as they finish where they start. <<<<<<<< are you sure?
The question then asks me to prove that axb = bxc = cxa. This seems obvious as if the values are the same then the cross products will be too. However i need to prove this and am not sure how to. Any help would be appreciated.

If the vector sum equals zero you have a closed curve.

Your 3 vectors form a triangle which determines allways a plane. The cross-product of such 2 vectors is the normal vector of the plane.
• Jan 6th 2010, 08:51 AM
Soroban
Hello, schteve!

Quote:

I have 3 vectors $\vec a, \vec b, \vec c$ such that: . $a+b+c\:=\:0$

This implies they are all the same length . . . . not true
and they finish where they start.

The question then asks me to prove that: . $\vec a\times\vec b \:=\: \vec b\times\vec c \:=\: \vec c\times\vec a$

Earboth showed you why the three cross products have the same direction.
Now we must prove they have the same magnitude (length).

The three vectors form a triangle.
Code:

            C             *           *  *       |b| *    *  |a|         *        *         *          *       *              *     A *  *  *  *  *  *  * B               |c|
Law of Sines: . $\frac{|a|}{\sin A} \:=\:\frac{|b|}{\sin B} \quad\Rightarrow\quad|a|\sin B \:=\:|b|\sin A$

Multiply both sides by $|c|\!:\;\;|a||c|\sin B \:=\:|b||c|\sin A$ .[1]

Recall that the magnitude of a cross product is the area
. . of the parallelogram determined by the two vectors.
Code:

          * - - - - - - *         /:::::::::::::/       v /:::::::::::::/       /:θ:::::::::::/       * - - - - - - *             u
That is: . $\text{Area} \:=\:|\vec u \times \vec v| \:=\:|u||v|\sin\theta$

Hence: . $\begin{array}{ccc}|\vec a \times \vec c| &=& |a||c|\sin B \\ \\[-4mm] |\vec b \times \vec c| &=& |b||c|sin A \end{array}$

According to [1], these two magnitudes are equal: . $|\vec a \times \vec c| \:=\:|\vec b \times \vec c|$

In a similar fashion, we can prove that all three magnitudes are equal.

• Jan 6th 2010, 11:49 AM
schteve
Thank you to both of the above respondents - it seems obvious now and i feel better for now knowing the answer but also slightly more stupid at the same time because i didn't get it straight away.