Coordinates of a point in a three dimensional picture
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Is there any way of working out the coordinates at $D$ in the following picture? $AD$ is perpendicular to $BD$ in the plane $ABC$
geometry trigonometry vectors
edited Dec 27 '18 at 1:27
Andrei
12k21126
asked Dec 26 '18 at 23:11
Alex CookeAlex Cooke
61
$endgroup$
add a comment |
$begingroup$
Is there any way of working out the coordinates at $D$ in the following picture? $AD$ is perpendicular to $BD$ in the plane $ABC$
geometry trigonometry vectors
edited Dec 27 '18 at 1:27
Andrei
12k21126
asked Dec 26 '18 at 23:11
Alex CookeAlex Cooke
61
$endgroup$
$begingroup$
What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
$endgroup$
– Andrei
Dec 27 '18 at 0:06
$begingroup$
AD⊥BD is correct, that's what I meant. Sorry for the confusion
$endgroup$
– Alex Cooke
Dec 27 '18 at 0:22
$begingroup$
There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
$endgroup$
– Dylan
Dec 27 '18 at 7:08
add a comment |
$begingroup$
Is there any way of working out the coordinates at $D$ in the following picture? $AD$ is perpendicular to $BD$ in the plane $ABC$
geometry trigonometry vectors
edited Dec 27 '18 at 1:27
Andrei
12k21126
asked Dec 26 '18 at 23:11
Alex CookeAlex Cooke
61
$endgroup$
Is there any way of working out the coordinates at $D$ in the following picture? $AD$ is perpendicular to $BD$ in the plane $ABC$
geometry trigonometry vectors
geometry trigonometry vectors
edited Dec 27 '18 at 1:27
Andrei
12k21126
asked Dec 26 '18 at 23:11
Alex CookeAlex Cooke
61
edited Dec 27 '18 at 1:27
Andrei
12k21126
asked Dec 26 '18 at 23:11
Alex CookeAlex Cooke
61
edited Dec 27 '18 at 1:27
Andrei
12k21126
edited Dec 27 '18 at 1:27
Andrei
12k21126
edited Dec 27 '18 at 1:27
Andrei
12k21126
12k21126
asked Dec 26 '18 at 23:11
Alex CookeAlex Cooke
61
asked Dec 26 '18 at 23:11
Alex CookeAlex Cooke
61
asked Dec 26 '18 at 23:11
Alex CookeAlex Cooke
61
61
$begingroup$
What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
$endgroup$
– Andrei
Dec 27 '18 at 0:06
$begingroup$
AD⊥BD is correct, that's what I meant. Sorry for the confusion
$endgroup$
– Alex Cooke
Dec 27 '18 at 0:22
$begingroup$
There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
$endgroup$
– Dylan
Dec 27 '18 at 7:08
add a comment |
$begingroup$
What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
$endgroup$
– Andrei
Dec 27 '18 at 0:06
$begingroup$
AD⊥BD is correct, that's what I meant. Sorry for the confusion
$endgroup$
– Alex Cooke
Dec 27 '18 at 0:22
$begingroup$
There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
$endgroup$
– Dylan
Dec 27 '18 at 7:08
$begingroup$
What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
$endgroup$
– Andrei
Dec 27 '18 at 0:06
$begingroup$
What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
$endgroup$
– Andrei
Dec 27 '18 at 0:06
$begingroup$
AD⊥BD is correct, that's what I meant. Sorry for the confusion
$endgroup$
– Alex Cooke
Dec 27 '18 at 0:22
$begingroup$
AD⊥BD is correct, that's what I meant. Sorry for the confusion
$endgroup$
– Alex Cooke
Dec 27 '18 at 0:22
$begingroup$
There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
$endgroup$
– Dylan
Dec 27 '18 at 7:08
$begingroup$
There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
$endgroup$
– Dylan
Dec 27 '18 at 7:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 27 '18 at 1:35
AndreiAndrei
12k21126
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Dec 27 '18 at 1:35
AndreiAndrei
12k21126
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 27 '18 at 1:35
AndreiAndrei
12k21126
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 27 '18 at 1:35
AndreiAndrei
12k21126
$endgroup$
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.
answered Dec 27 '18 at 1:35
AndreiAndrei
12k21126
answered Dec 27 '18 at 1:35
AndreiAndrei
12k21126
answered Dec 27 '18 at 1:35
AndreiAndrei
12k21126
answered Dec 27 '18 at 1:35
AndreiAndrei
12k21126
12k21126
add a comment |
add a comment |
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$begingroup$
What does it mean that a point is perpendicular to a line? And from the figure it looks like $ADperp BD$
$endgroup$
– Andrei
Dec 27 '18 at 0:06
$begingroup$
AD⊥BD is correct, that's what I meant. Sorry for the confusion
$endgroup$
– Alex Cooke
Dec 27 '18 at 0:22
$begingroup$
There's not enough information. D can be any point on the semicircle with diameter AB on the plane ABC.
$endgroup$
– Dylan
Dec 27 '18 at 7:08