Find the equation in the form of ax+by+cz=d

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SUMMARY

The discussion focuses on solving two mathematical problems: finding the distance from a point P=(-3,4,2) to a line L defined by the parametric equation L:(x,y,z)=(3,-2,-1)+(1,-2,2)t, and deriving the equation of a plane in the form ax+by+cz=d that passes through point P and is parallel to line L. The distance is determined by minimizing the angle between the vector from point P to any point q on line L and the direction vector of line L, v=(1,-2,2). The equation of the plane can be established using the normal vector derived from the direction vector of line L.

PREREQUISITES
  • Understanding of vector operations and geometry in three-dimensional space.
  • Familiarity with parametric equations of lines.
  • Knowledge of the concept of distance between a point and a line.
  • Ability to derive equations of planes from points and direction vectors.
NEXT STEPS
  • Study vector projections to understand how to minimize distances in three-dimensional space.
  • Learn about the derivation of plane equations from points and direction vectors.
  • Explore the concept of normal vectors and their role in defining planes.
  • Practice solving similar problems involving distances from points to lines and planes in 3D geometry.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with three-dimensional geometry and need to solve problems related to distances and plane equations.

XBOX999
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Hello,
I need help in this two quetion. Please, show me all the steps to solve them.

1- Find the distence from a point P= ( -3,4,2) to a line L:(x,y,z)= ( 3,-2,-1)+(1,-2,2)t.



2-find the equation in the form of ax+by+cz=d, of the plane going through a point
P= (-3,4,2) to a line L:(x,y,z)= (3,-2,-1) + (1,-2,2)t.
 
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i'm at the library, i was just looking at those in my crc standard formulas and tables around page 240-250. its a piece of cake but i can't quote them. someone will show up.
 
XBOX999 said:
Hello,
I need help in this two quetion. Please, show me all the steps to solve them.

1- Find the distence from a point P= ( -3,4,2) to a line L:(x,y,z)= ( 3,-2,-1)+(1,-2,2)t.
2-find the equation in the form of ax+by+cz=d, of the plane going through a point
P= (-3,4,2) to a line L:(x,y,z)= (3,-2,-1) + (1,-2,2)t.

For the first part, the distance from a point to a line is by definition the SHORTEST possible distance along a line connecting P and L. At what angle must the two lines meet in order for the distance to be minimized?

In other words, if q is a point on the line, you want ||q - p|| to be the shortest possible. If v is a vector pointing in the direction of the line, what condition must the vectors (q - p) and v satisfy?
 

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