FInd the Vector passing thru a point and parallel to two points in a plane

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SUMMARY

This discussion focuses on finding a vector that passes through point B' and is parallel to the line defined by points A' and C' in a three-dimensional space. The points are defined as A'=(9,27,-0.6), B'=(12,27,-6), and C'=(19,25,-8). The user initially calculated the vectors A'B' and A'C', then found the normal vector to the plane containing these points using the cross product. The equation of the plane was derived as -48x + 30y - 2z = -90. The final goal is to express the line through B' in the direction of the vector A'C'.

PREREQUISITES
  • Understanding of vector operations, including addition and subtraction.
  • Familiarity with the cross product and its geometric interpretation.
  • Knowledge of parametric equations of lines in three-dimensional space.
  • Ability to derive equations of planes from points and vectors.
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  • Study the properties of cross products in vector mathematics.
  • Learn how to derive parametric equations for lines in 3D space.
  • Explore methods for finding normal vectors to planes defined by three points.
  • Investigate applications of vectors in computer graphics and physics simulations.
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Mathematicians, physics students, computer graphics developers, and anyone involved in 3D modeling or spatial analysis will benefit from this discussion.

kailsen
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Hi,
i have three points A',B',C' and want to find a vector passing through B' and parallel to the other two points in the plane containing these 3 points.
So this how i started,

A'=(9,27,-0.6)
B'=(12,27,-6)
C'=(19,25,-8)

I found the vector AB' and AC' first.
A'B'= A'-B'
A'C'=A'-C'

Found cross product of the two resultant vector A'B' and A'C' to get a normal vector.
now using the equation a*[x – xA] + b*[y – yA] + c*[z – zA] = 0,
i found the equation of the plane that contains all these 3 vectors (A',B',C').
Assuming the equation is -48x + 30y -2z = -90.
Now i would like to get a vector (basically x,y,z) that is passing through B' and parallel to the line passing through A' and C'.

I think, finding a line passing through the origin of of the normal vector to the plane will fetch me half the answer, but how to satisfy it to be parallel to the line A'C'
 
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First, you statement of the problem is strange. A "vector" does not pass through a particular point. It is entirely given by its direction and length. Nor is one vector "parallel" to another though they may be "in the same direction. You may mean "find the equation of the line passing through B' parallel to the line through A' and C'".

But then I have no idea why you would go to the work of finding two distinct vectors and their dot product. The vector from A' to C' is <19- 9, 25- 27, -8- (-0.6)>= <10, -2, -7.4>. The line through B'= (12, 27, -6) in the direction of that vector is given by
x= 10t+ 12, y= -2t+ 27, z= -7.4t- 6.
 
kailsen said:
Hi,
i have three points A',B',C' and want to find a vector passing through B' and parallel to the other two points in the plane containing these 3 points.
So this how i started,

A'=(9,27,-0.6)
B'=(12,27,-6)
C'=(19,25,-8)

I found the vector AB' and AC' first.
A'B'= A'-B'
A'C'=A'-C'

Found cross product of the two resultant vector A'B' and A'C' to get a normal vector.
now using the equation a*[x – xA] + b*[y – yA] + c*[z – zA] = 0,
i found the equation of the plane that contains all these 3 vectors (A',B',C').
Assuming the equation is -48x + 30y -2z = -90.
Now i would like to get a vector (basically x,y,z) that is passing through B' and parallel to the line passing through A' and C'.

I think, finding a line passing through the origin of of the normal vector to the plane will fetch me half the answer, but how to satisfy it to be parallel to the line A'C'

The line you want can be expressed in parametric form (s is parameter) as B' + s(A'-C').
 

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