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