Vectors Cartesian equations and normals

  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
Simon Bridge
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Define "cartesian form".
 
  • #3
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  • #4
Simon Bridge
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so f(x)=g(y)=h(z) means f-g-h=0 ...
 
  • #5
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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?
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.
 
  • #6
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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.
 
  • #7
Simon Bridge
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Yeah - the idea is to get OP to talk about what it means and so work out a method.
 
  • #8
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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
 
  • #9
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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
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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
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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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