Vectors Cartesian equations and normals

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    Cartesian Vectors
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Discussion Overview

The discussion revolves around finding the Cartesian equation of a line that passes through a specific point and is normal to a given plane equation. Participants explore the definitions and forms of Cartesian equations, as well as the relationship between lines and planes in three-dimensional space.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant describes their method for finding the Cartesian equation of the line normal to the plane defined by -3x-2y+2z=0, using the normal direction derived from the plane's equation.
  • Several participants seek clarification on what constitutes "Cartesian form," suggesting different interpretations and representations of equations for lines.
  • Another participant proposes a method for expressing the line in Cartesian form, referencing a general equation involving a point and a direction vector.
  • There is a discussion about whether the derived equations satisfy the original problem statement, specifically regarding the perpendicularity of the line to the plane and the inclusion of the specified point.
  • One participant questions the necessity of confirming perpendicularity, while another clarifies that a line can be normal to a plane, which is synonymous with being perpendicular.
  • Participants discuss the use of the dot product to verify perpendicularity, with one suggesting that it may not be necessary given the normal vector was chosen from the plane's equation.
  • There is mention of an alternative representation of the plane's equation, which leads to a discussion about the equivalence of different normal vectors.

Areas of Agreement / Disagreement

Participants express differing views on the definitions and forms of Cartesian equations, as well as the necessity of verifying perpendicularity. The discussion remains unresolved regarding the correctness of the initial method and the interpretation of Cartesian form.

Contextual Notes

Some participants highlight the ambiguity in defining "Cartesian form" and the implications of different representations. There is also an acknowledgment of the potential confusion surrounding the relationship between lines and planes.

Who May Find This Useful

This discussion may be useful for students and individuals interested in understanding the geometric relationships between lines and planes, as well as those seeking clarification on Cartesian equations in three-dimensional space.

53Mark53
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I am having trouble finding the Cartesian equation of the line k which passes through Y(0,1,0) and is normal to -3x-2y+2z=0

this is what I tried to do but not sure if it is the correct method

normal direction (-3,-2,2)

-3x-2y+2z=d

sub in (0,1,0) to -3x-2y+2z=d

d=-2

-3x-2y+2z=-2

is this correct if so how would i get it into cartesian form?
 
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Define "cartesian form".
 
Simon Bridge said:
Define "cartesian form".
3x=2y=1-z for example
 
so f(x)=g(y)=h(z) means f-g-h=0 ...
 
53Mark53 said:
I am having trouble finding the Cartesian equation of the line k which passes through Y(0,1,0) and is normal to -3x-2y+2z=0

this is what I tried to do but not sure if it is the correct method

normal direction (-3,-2,2)

-3x-2y+2z=d

sub in (0,1,0) to -3x-2y+2z=d

d=-2

-3x-2y+2z=-2

is this correct if so how would i get it into cartesian form?
If line L is parallel to the vector v = <A, B, C> and contains the point ##P(x_0, y_0, z_0)##, its equation in Cartesian form is
$$\frac{x - x_0}{A} = \frac{y - y_0}{B} = \frac{z - z_0}{C}$$

In cases where one or more of the coordinates of v happens to be zero, it's permitted to write a denominator of 0 in the associated fraction.
 
Simon Bridge said:
Define "cartesian form".
This is a well-known term for writing equations for a line. Another possibility is parametric form; i.e., x = f(t) + x0, y = g(t) + y0, z = h(t) + z0.
 
Yeah - the idea is to get OP to talk about what it means and so work out a method.
 
Mark44 said:
If line L is parallel to the vector v = <A, B, C> and contains the point ##P(x_0, y_0, z_0)##, its equation in Cartesian form is
$$\frac{x - x_0}{A} = \frac{y - y_0}{B} = \frac{z - z_0}{C}$$

In cases where one or more of the coordinates of v happens to be zero, it's permitted to write a denominator of 0 in the associated fraction.

does this mean the answer would be -x/3=-(y-1)/2=z/2
 
53Mark53 said:
does this mean the answer would be -x/3=-(y-1)/2=z/2
Do these equations satisfy the problem statement? IOW, is the direction of this line perpendicular to the given plane, and is the given point on this line?

BTW, homework questions should be posted in the Homework & Coursework sections, not here in the technical math sections.
 
  • #10
Mark44 said:
Do these equations satisfy the problem statement? IOW, is the direction of this line perpendicular to the given plane, and is the given point on this line?

BTW, homework questions should be posted in the Homework & Coursework sections, not here in the technical math sections.
I am trying to find the line that is normal to the equation why do I need know if it perpendicular to the plane?

would I use the dot product to see if it is perpendicular?
 
  • #11
53Mark53 said:
I am trying to find the line that is normal to the equation why do I need know if it perpendicular to the plane?
A line isn't "normal to an equation." It can be normal to a plane, which is exactly the same as saying the line is perpendicular to the plane.
53Mark53 said:
would I use the dot product to see if it is perpendicular?
Sure you could do that, but it's not necessary. You chose the vector <-3, -2, 2> by inspection, I think, from the plane's equation -3x - 2y + 2z = 0. The line you found has the same direction as the plane's normal, right?

Since you're uncertain about things, maybe verifying that the normal is perpendicular to the plane is a good idea. To do this, take the dot product of any vector that lies in the plane with the normal vector. You can find a vector in the plane by using the plane's equation to find two points, and forming a vector between these two points. Then take the dot product of that vector and the normal <-3, -2, 2>.

BTW, the equation of the plane could also be written as 3x + 2y - 2z = 0, with the normal being <3, 2, -2>. The two equations are equivalent, meaning that they both describe exactly the same set of points, but the vector <3, 2, -2> points in the opposite direction as the normal you found. Although these two vectors are different, both are perpendicular to the plane in this problem.
 

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