- #1

susmit

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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.

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- Thread starter susmit
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- #1

susmit

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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

Pyrrhus

Homework Helper

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Set up your system of equations.

[tex] |\vec{a}| \hat{u_{1}} + |\vec{b}| \hat{u_{1}} = 20\hat{u_{1}} [/tex]

[tex] |\vec{a}| \hat{u_{1}} + |\vec{b}| \hat{u_{2}} = 10 \hat{u_{3}} [/tex]

also

[tex] \hat{u_{3}} = \cos \theta \hat{u_{1}} + \sin \theta \hat{u_{2}} [/tex]

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].

[tex] |\vec{a}| \hat{u_{1}} + |\vec{b}| \hat{u_{1}} = 20\hat{u_{1}} [/tex]

[tex] |\vec{a}| \hat{u_{1}} + |\vec{b}| \hat{u_{2}} = 10 \hat{u_{3}} [/tex]

also

[tex] \hat{u_{3}} = \cos \theta \hat{u_{1}} + \sin \theta \hat{u_{2}} [/tex]

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].

Last edited:

- #3

Meir Achuz

Science Advisor

Homework Helper

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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

Antiphon

- 1,683

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Meir Achuz said:

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.

- #5

Crosson

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- #6

Antiphon

- 1,683

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

[tex] \vec{v1} = \hat{x} [ \sqrt{50} (\sqrt{2} + i) ] [/tex]

[tex] \vec{v2} = \hat{a} [ \sqrt{50} (\sqrt{2} - i) ] [/tex]

[tex] \hat{a} = \hat{x} [/tex] in the parallel case and

[tex] \hat{a} = \hat{y} [/tex] in the perpendicular case.

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

[tex] \vec{v1} = \hat{x} [ \sqrt{50} (\sqrt{2} + i) ] [/tex]

[tex] \vec{v2} = \hat{a} [ \sqrt{50} (\sqrt{2} - i) ] [/tex]

[tex] \hat{a} = \hat{x} [/tex] in the parallel case and

[tex] \hat{a} = \hat{y} [/tex] in the perpendicular case.

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

Last edited:

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