# Imaginary vectors

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.

Pyrrhus
Homework Helper
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 $\hat{u_{3}}$.

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Meir Achuz
Homework Helper
Gold Member
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.

Meir Achuz said:
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.

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

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.

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.

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