# Imaginary vectors

1. Jul 25, 2005

### susmit

hi
here's a questn:
"2 vectors acting in same direction have resultant 20 whereas in perpendicular direction resultant is 10. find the vectors."

pls. explain the imaginary solution.

2. Jul 25, 2005

### Pyrrhus

Set up your system of equations.

$$|\vec{a}| \hat{u_{1}} + |\vec{b}| \hat{u_{1}} = 20\hat{u_{1}}$$

$$|\vec{a}| \hat{u_{1}} + |\vec{b}| \hat{u_{2}} = 10 \hat{u_{3}}$$

also

$$\hat{u_{3}} = \cos \theta \hat{u_{1}} + \sin \theta \hat{u_{2}}$$

where $\hat{u_{1}}$, $\hat{u_{2}}$, $\hat{u_{3}}$ are unit vectors, and $\theta$ is the angle between $\hat{u_{1}}$ and $\hat{u_{3}}$.

Last edited: Jul 25, 2005
3. Jul 25, 2005

### Meir Achuz

In your case, the two equations are
a+b=20
a^2+b^2=100.
This leads to a quadratic equation with (as you say) no real solution.
This means that you have described an impossible situation.

4. Jul 25, 2005

### Antiphon

Not impossible, just imaginary. The solution lies in the field
of complex numbers i.e. complex vectors.

5. Jul 25, 2005

### Crosson

This equivalent to "construct a right triangle so that the sum of legs is twice that of the hypotenuse". This is equivalent to the equation a + b = 2c which would suggest that a^2 + b^2 =4c^2-2ab. The only way this can be consistent (given the pythagorean relation is if) c^2=4c^2 - 2ab or ab =3/2 c^2 which is a violation of the CS inequality, so there is no solution.

6. Jul 25, 2005

### Antiphon

I didn't want to solve this kid's problem, Crosson but you forced me to.

$$\vec{v1} = \hat{x} [ \sqrt{50} (\sqrt{2} + i) ]$$
$$\vec{v2} = \hat{a} [ \sqrt{50} (\sqrt{2} - i) ]$$

$$\hat{a} = \hat{x}$$ in the parallel case and
$$\hat{a} = \hat{y}$$ in the perpendicular case.

Edit: The solution is not unique, I just picked one.

Last edited: Jul 25, 2005