Coordinates of a point in a three dimensional picture












-1















Is there any way of working out the coordinates at $D$ in the following picture? $AD$ is perpendicular to $BD$ in the plane $ABC$




Link to picture showing coordinates,lengths and angles










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  • What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
    – Andrei
    Dec 27 '18 at 0:06










  • AD⊥BD is correct, that's what I meant. Sorry for the confusion
    – Alex Cooke
    Dec 27 '18 at 0:22












  • There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
    – Dylan
    Dec 27 '18 at 7:08
















-1















Is there any way of working out the coordinates at $D$ in the following picture? $AD$ is perpendicular to $BD$ in the plane $ABC$




Link to picture showing coordinates,lengths and angles










share|cite|improve this question









New contributor




Alex Cooke is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
    – Andrei
    Dec 27 '18 at 0:06










  • AD⊥BD is correct, that's what I meant. Sorry for the confusion
    – Alex Cooke
    Dec 27 '18 at 0:22












  • There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
    – Dylan
    Dec 27 '18 at 7:08














-1












-1








-1








Is there any way of working out the coordinates at $D$ in the following picture? $AD$ is perpendicular to $BD$ in the plane $ABC$




Link to picture showing coordinates,lengths and angles










share|cite|improve this question









New contributor




Alex Cooke is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Is there any way of working out the coordinates at $D$ in the following picture? $AD$ is perpendicular to $BD$ in the plane $ABC$




Link to picture showing coordinates,lengths and angles







geometry trigonometry vectors






share|cite|improve this question









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Alex Cooke is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question









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Alex Cooke is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited Dec 27 '18 at 1:27









Andrei

11.2k21026




11.2k21026






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asked Dec 26 '18 at 23:11









Alex Cooke

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New contributor





Alex Cooke is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Alex Cooke is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
    – Andrei
    Dec 27 '18 at 0:06










  • AD⊥BD is correct, that's what I meant. Sorry for the confusion
    – Alex Cooke
    Dec 27 '18 at 0:22












  • There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
    – Dylan
    Dec 27 '18 at 7:08


















  • What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
    – Andrei
    Dec 27 '18 at 0:06










  • AD⊥BD is correct, that's what I meant. Sorry for the confusion
    – Alex Cooke
    Dec 27 '18 at 0:22












  • There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
    – Dylan
    Dec 27 '18 at 7:08
















What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
– Andrei
Dec 27 '18 at 0:06




What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
– Andrei
Dec 27 '18 at 0:06












AD⊥BD is correct, that's what I meant. Sorry for the confusion
– Alex Cooke
Dec 27 '18 at 0:22






AD⊥BD is correct, that's what I meant. Sorry for the confusion
– Alex Cooke
Dec 27 '18 at 0:22














There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
– Dylan
Dec 27 '18 at 7:08




There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
– Dylan
Dec 27 '18 at 7:08










1 Answer
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There is no unique $D$ point. The locus of such points is the circle in the $ABC$ plane with $AB$ as diameter. If you put the condition that $AD=BD$, then you are down to two points, one one the same side of $AB$ as $C$, one exactly opposite.



To construct such points, find the center of $AB$, and the radius (length of $AB$ divided by $2$. Then write the equation of the sphere with this radius and center, and the equation of the $ABC$ plane. Any point $D$ in the intersection verifies your first condition. To have $AD=BD$, you also add the equation of the plane perpendicular to $AB$ through the center, and solve the system of equations. You should get two solutions.






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    1 Answer
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    1 Answer
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    There is no unique $D$ point. The locus of such points is the circle in the $ABC$ plane with $AB$ as diameter. If you put the condition that $AD=BD$, then you are down to two points, one one the same side of $AB$ as $C$, one exactly opposite.



    To construct such points, find the center of $AB$, and the radius (length of $AB$ divided by $2$. Then write the equation of the sphere with this radius and center, and the equation of the $ABC$ plane. Any point $D$ in the intersection verifies your first condition. To have $AD=BD$, you also add the equation of the plane perpendicular to $AB$ through the center, and solve the system of equations. You should get two solutions.






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      There is no unique $D$ point. The locus of such points is the circle in the $ABC$ plane with $AB$ as diameter. If you put the condition that $AD=BD$, then you are down to two points, one one the same side of $AB$ as $C$, one exactly opposite.



      To construct such points, find the center of $AB$, and the radius (length of $AB$ divided by $2$. Then write the equation of the sphere with this radius and center, and the equation of the $ABC$ plane. Any point $D$ in the intersection verifies your first condition. To have $AD=BD$, you also add the equation of the plane perpendicular to $AB$ through the center, and solve the system of equations. You should get two solutions.






      share|cite|improve this answer
























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        0






        There is no unique $D$ point. The locus of such points is the circle in the $ABC$ plane with $AB$ as diameter. If you put the condition that $AD=BD$, then you are down to two points, one one the same side of $AB$ as $C$, one exactly opposite.



        To construct such points, find the center of $AB$, and the radius (length of $AB$ divided by $2$. Then write the equation of the sphere with this radius and center, and the equation of the $ABC$ plane. Any point $D$ in the intersection verifies your first condition. To have $AD=BD$, you also add the equation of the plane perpendicular to $AB$ through the center, and solve the system of equations. You should get two solutions.






        share|cite|improve this answer












        There is no unique $D$ point. The locus of such points is the circle in the $ABC$ plane with $AB$ as diameter. If you put the condition that $AD=BD$, then you are down to two points, one one the same side of $AB$ as $C$, one exactly opposite.



        To construct such points, find the center of $AB$, and the radius (length of $AB$ divided by $2$. Then write the equation of the sphere with this radius and center, and the equation of the $ABC$ plane. Any point $D$ in the intersection verifies your first condition. To have $AD=BD$, you also add the equation of the plane perpendicular to $AB$ through the center, and solve the system of equations. You should get two solutions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 1:35









        Andrei

        11.2k21026




        11.2k21026






















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