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Question Number 2092 by Filup last updated on 02/Nov/15
If A=⟨p_x , p_y , p_z ⟩ is a position vector   in standard position, vector r=⟨r_x , r_y , r_z ⟩  is the radius of a 3 dimensional circle  with focal point at A, Solve for the  vector equation r(θ) such that it is the  equation of a 3 dimensional circle
$$\mathrm{If}\:\boldsymbol{{A}}=\langle{p}_{{x}} ,\:{p}_{{y}} ,\:{p}_{{z}} \rangle\:\mathrm{is}\:\mathrm{a}\:\mathrm{position}\:\mathrm{vector}\: \\ $$$$\mathrm{in}\:\mathrm{standard}\:\mathrm{position},\:\mathrm{vector}\:\boldsymbol{{r}}=\langle{r}_{{x}} ,\:{r}_{{y}} ,\:{r}_{{z}} \rangle \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{a}\:\mathrm{3}\:\mathrm{dimensional}\:\mathrm{circle} \\ $$$$\mathrm{with}\:\mathrm{focal}\:\mathrm{point}\:\mathrm{at}\:\boldsymbol{{A}},\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{vector}\:\mathrm{equation}\:\boldsymbol{{r}}\left(\theta\right)\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{3}\:\mathrm{dimensional}\:\mathrm{circle} \\ $$
Commented by Filup last updated on 02/Nov/15
Solve such that it is in form:  r(θ)=⟨x(θ), y(θ), z(θ)⟩
$$\mathrm{Solve}\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is}\:\mathrm{in}\:\mathrm{form}: \\ $$$$\boldsymbol{{r}}\left(\theta\right)=\langle{x}\left(\theta\right),\:{y}\left(\theta\right),\:{z}\left(\theta\right)\rangle \\ $$
Commented by prakash jain last updated on 03/Nov/15
You have a point and radius vector in 3D.   There will be many circles possible.
$$\mathrm{You}\:\mathrm{have}\:\mathrm{a}\:\mathrm{point}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{vector}\:\mathrm{in}\:\mathrm{3D}.\: \\ $$$$\mathrm{There}\:\mathrm{will}\:\mathrm{be}\:\mathrm{many}\:\mathrm{circles}\:\mathrm{possible}. \\ $$

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