Thread: Finding a vector given another vector and their cross product

Hello,

I seem to have forgotten how to solve this type of problem as even by constructing vectors from random number, getting their cross product, substituting one vector with dummy variable and attempting to solve for the vector, I keep getting trivial solutions. Can someone please explain what I'm doing wrong. Here is an example:

U = 1i +2j +3k
V = 4i +5j +6k

UxV = (12-15)i - (6-12)j + (5-8)k = -3i +6j -3k

Now use V and UxV to find U:

let U = ai + bj + ck

UxV = (6b - 5c)i - (6a - 4c)j + (5a - 4b)k

Gives the following system:

6b - 5c = -3 (1)
-6a + 4c = 6 (2)
5a - 4b = -3 (3)

From (1): b = 5c/6 - 1/2
from (2): a = 2c/3 - 1
Plugging into (3): 10c/3 - 5 - 10c/3 +2 = -3
-3 = -3

I run into something similar whether its a made up problem or one from a text that's designed to be solveable. Any help is much appreciated.

J

Originally Posted by mathj

U = 1i +2j +3k
V = 4i +5j +6k
UxV = (12-15)i - (6-12)j + (5-8)k = -3i +6j -3k CORRECT

Now use V and UxV to find U:

Pray tell, why in the world would you be finding U when you already know it? That do not make sense.

I am using this as an exercise to practice solving this type of problem. Finding a vector that I already know is confirmation that have arrived at the correct (and desired, when multiple answers exist) solution. In this case, doing so has illustrated to me the fact that I have some problem in my technique or reasoning. If a problem without a known solution will make this more clear for you, a text book problem of this type follows.

U x V = -30i + 40k

V = 4i - 2j + 3k

Find U

Again, with this I still run into a trivial solution, though an expected solution exists.

I hope this sheds some light on my motivation for this. I appreciate any help you can offer.

J

Originally Posted by mathj

I am using this as an exercise to practice solving this type of problem.
U x V = -30i + 40k
V = 4i - 2j + 3
Find U

If then
You also know that
Do you expect to find a unique solution?

Hello, mathj!

Plato is absolutely correct.
Your "trivial solution" indicates that there is an infinite number of solutions.
I'll explain this with algebra.

.

. .

Then we have the system: .



And we have: .
. . which is identical to equation [1].


So we have: .

. . From [3]: .

Hence, we have: .

On the right, replace with the parameter .

Therefore: .

The text seemed to imply there was a specific solution sought. However, using soroban's final set of eqns I can see where the back-of-the-book soln comes from.

soroban, your respons was just what I was looking for. Thanks very much for the detail.

J