Finding points on a line in 3d

In summary, the conversation is about finding points algebraically given three points in 3D space with equal distances between them. The attempt at a solution involves using vector representation and finding the length of each vector. The equations for the length of AB, BC, and AC are given and can be used to solve for the points algebraically.
  • #1
1
0

Homework Statement


It's been a while since I've done this stuff D:

So Given Points A(a,1,0), B(1,b,0) and C(1,0,c) with |AC|=|BC|=|AB|

Find the points algebraically


Homework Equations



Unfortunately, i don't have much knowledge on this. I am sure there's a 3d point slope style formula around somewhere though :s


The Attempt at a Solution



At first my guess was to try making it a vector such that SQRT((Xc-Xa)^2+(Yc-Ya)^2+(Zc-Za)^2) and so on, but from there i was confused on how to find the points.

I appreciate the potential help, thanks guys :D
 
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  • #2
Mosier said:

Homework Statement


It's been a while since I've done this stuff D:

So Given Points A(a,1,0), B(1,b,0) and C(1,0,c) with |AC|=|BC|=|AB|

Find the points algebraically


Homework Equations



Unfortunately, i don't have much knowledge on this. I am sure there's a 3d point slope style formula around somewhere though :s


The Attempt at a Solution



At first my guess was to try making it a vector such that SQRT((Xc-Xa)^2+(Yc-Ya)^2+(Zc-Za)^2) and so on, but from there i was confused on how to find the points.

I appreciate the potential help, thanks guys :D

Is there a need for X,Y and Z to be in the square? :smile:
 
  • #3
Mosier said:

Homework Statement


It's been a while since I've done this stuff D:

So Given Points A(a,1,0), B(1,b,0) and C(1,0,c) with |AC|=|BC|=|AB|

Find the points algebraically


Homework Equations



Unfortunately, i don't have much knowledge on this. I am sure there's a 3d point slope style formula around somewhere though :s


The Attempt at a Solution



At first my guess was to try making it a vector such that SQRT((Xc-Xa)^2+(Yc-Ya)^2+(Zc-Za)^2) and so on, but from there i was confused on how to find the points.

I appreciate the potential help, thanks guys :D

Write the vector that represents the vector AB. Write an expression that represents its length. For example, AB = (1 - a, b - 1, 0). |AB| = sqrt((1 - a)^2 + (b - 1)^2).
Do the same with vector BC. Write an expression that represents its length.
Do the same with vector AC. Write an expression that represents its length.

You are given that |AB| = |BC|, |BC| = |AC|, and |AB| = |BC|. Substitute the expressions you already found in these three equations.
 

1. How do you find a point on a line in 3D?

To find a point on a line in 3D, you need at least one known point on the line and the direction of the line. You can then use the point-slope formula or parametric equations to find the coordinates of the desired point.

2. What is the point-slope formula for finding a point on a line in 3D?

The point-slope formula for finding a point on a line in 3D is (x,y,z) = (x1, y1, z1) + t(a,b,c), where (x1, y1, z1) is a known point on the line and (a,b,c) is the direction vector of the line.

3. What are parametric equations and how are they used to find points on a line in 3D?

Parametric equations are equations that describe the coordinates of a point on a line in terms of a parameter. In 3D, they can be written as x = x1 + at, y = y1 + bt, and z = z1 + ct, where (x1, y1, z1) is a known point on the line, (a,b,c) is the direction vector, and t is the parameter.

4. Can you find a point on a line in 3D with only two known points?

No, you need at least one known point and the direction of the line in order to find a point on a line in 3D. With only two known points, you can determine the direction of the line, but not the specific coordinates of a point on it.

5. How do you determine if a point is on a line in 3D?

A point (x,y,z) is on a line if it satisfies the equation (x,y,z) = (x1, y1, z1) + t(a,b,c), where (x1, y1, z1) is a known point on the line, (a,b,c) is the direction vector, and t is the parameter. If the coordinates of the point satisfy this equation, then it is on the line.

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