Finding point-normal equations

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In summary: Thanks for the help LCKurtz!In summary, the point-normal equation for the line L is obtained by eliminating the parameter k from the given equation and setting the equations for x and y equal to each other. This results in a form of (for some constants a and b and some points x0 and y0 on the line L): a(x-x0)+b(y-y0)=0, where a and b are the normal to the line L.
  • #1
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Homework Statement


Say that I have a problem along the lines of: Find the point-normal equation for the line L.

L=u1+k*u2, where k is ℝ.

And then I'm given u1 and u2, which both are 2×1 vectors.

Then L will also be a 2×1 vector (which consists of a11 and a21), which I guess is ok (since it gives the two coordinats in R2).

However the point-normal equation is in the form of (for some constants a and b and some points x0 and y0 on the line L):

a(x-x0)+b(y-0)=0.

So here comes my question: In the vector L: Are a11 or a21 x?


Homework Equations


-

The Attempt at a Solution


See above.
 
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  • #2
Computnik said:

Homework Statement


Say that I have a problem along the lines of: Find the point-normal equation for the line L.

L=u1+k*u2, where k is ℝ.

And then I'm given u1 and u2, which both are 2×1 vectors.

Then L will also be a 2×1 vector (which consists of a11 and a21), which I guess is ok (since it gives the two coordinats in R2).

However the point-normal equation is in the form of (for some constants a and b and some points x0 and y0 on the line L):

a(x-x0)+b(y-0)=0.

So here comes my question: In the vector L: Are a11 or a21 x?


Homework Equations


-


The Attempt at a Solution


See above.

Let ##u_1=\langle a, b\rangle## and ##u_2=\langle d_1,d_2\rangle##. Then your line is ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle##. Eliminate the ##k## between the ##x## and ##y## equations and put it in your required form. Note that ##(a,b)## is a point on the line.
 
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  • #3
LCKurtz said:
Let ##u_1=\langle a, b\rangle## and ##u_2=\langle d_1,d_2\rangle##. Then your line is ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle##. Eliminate the ##k## between the ##x## and ##y## equations and put it in your required form. Note that ##(a,b)## is a point on the line.
Two questions:

1. (Perhaps a dumb question) Is ##\langle x,y\rangle## another way of writing a 2×1 matrix where x=a11 and y=a21? (I'm really new to linear algebra...)

2. If I'm correct about question 1, then I still don't see how I can "translate" ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle## to the form of

a(x-x0)+b(y-y0)=0

(That was what my original question was all about.)

Thanks for the help LCKurtz!
 
  • #4
Computnik said:
Two questions:

1. (Perhaps a dumb question) Is ##\langle x,y\rangle## another way of writing a 2×1 matrix where x=a11 and y=a21? (I'm really new to linear algebra...)

It does't matter whether you use a 2x1 or 1x2 matrix in this problem. You have two entries either way.

2. If I'm correct about question 1, then I still don't see how I can "translate" ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle## to the form of

a(x-x0)+b(y-y0)=0

(That was what my original question was all about.)

Thanks for the help LCKurtz!

Do what I suggested. Write the equations for ##x## and ##y## from the given matrix equation and eliminate the ##k## parameter. Try it.
 
  • #5
LCKurtz said:
It does't matter whether you use a 2x1 or 1x2 matrix in this problem. You have two entries either way.
Do what I suggested. Write the equations for ##x## and ##y## from the given matrix equation and eliminate the ##k## parameter. Try it.
I'm not sure what you mean with "eliminate the ##k## parameter".

But I guess this is a start?

##x=a+kd_1##

and

##y=b+kd_2##

now let me put ##x## and ##y## into the "form" of the point-normal equation. Which I will call:

##p(x-x_0)+q(y-y_0)=0##, where ##p## and ##q## are constants.

We know that ##x_0=a## and ##y_0=b## are solutions because ##k## could be ##0##.

so

##p((a+kd_1)-a)+q((b+kd_2)-b)=0##

##p(kd_1)+q(kd_2)=0##

##k(pd_1+qd_2)=0##

##pd_1+qd_2=0##

where ##(p,d)## is the normal to ##(x,y)##. But how do you get ##(p,d)##?

Does the calculations look ok LCKurtz?
 
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  • #6
Not quite. From ##x=a+kd_1,~y=b+kd_2## you eliminate ##k##. Solve each equation for ##k## and set them equal:$$
k = \frac {x-a}{d_1} = \frac {y-b}{d_2}$$Put that last equation in the form$$
(?)(x-a) + (??)(y-b) = 0$$and you will have your form.
 
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  • #7
I'm a bit lost. :/ Any more tips?
 
  • #8
LCKurtz said:
Not quite. From ##x=a+kd_1,~y=b+kd_2## you eliminate ##k##. Solve each equation for ##k## and set them equal:$$
k = \frac {x-a}{d_1} = \frac {y-b}{d_2}$$Put that last equation in the form$$
(?)(x-a) + (??)(y-b) = 0$$and you will have your form.
I know that the ##(?)## ##(??)## are the normal to the line ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle##. However what I don't know is how you should obtain the normal for that line...
 
  • #9
You say you are taking linear algebra but the problems you are having are all much more basic algebra matters.

[tex]\frac{x-a}{d_1}= \frac{y- b}{d_2}[/tex]

An obvious thing to do is to multiply both sides by [itex]d_1d_2[/itex] to get rid of the fractions:
[tex]d_2(x- a)= d_1(y- b)[/tex]
[tex]d_2(x- a)- d_1(y- b)= 0[/tex]
 
  • #10
HallsofIvy said:
You say you are taking linear algebra but the problems you are having are all much more basic algebra matters.

[tex]\frac{x-a}{d_1}= \frac{y- b}{d_2}[/tex]

An obvious thing to do is to multiply both sides by [itex]d_1d_2[/itex] to get rid of the fractions:
[tex]d_2(x- a)= d_1(y- b)[/tex]
[tex]d_2(x- a)- d_1(y- b)= 0[/tex]
I feel like such an idiot now, that was my initial thought but I dismissed it because it felt like it was too easy to be the right answer. :redface:
 

1. What is a point-normal equation?

A point-normal equation is a mathematical representation of a line or plane in three-dimensional space. It consists of a point on the line or plane and a normal vector that is perpendicular to the line or plane.

2. How do you find the point-normal equation of a line?

To find the point-normal equation of a line, you need to have a point on the line and a vector that is perpendicular to the line. You can then use the formula (x - x0) / a = (y - y0) / b = (z - z0) / c, where (x0, y0, z0) is the given point and (a, b, c) is the normal vector.

3. Can you find the point-normal equation of a plane?

Yes, you can find the point-normal equation of a plane. Similar to finding the equation of a line, you need a point on the plane and a normal vector. You can use the formula a(x - x0) + b(y - y0) + c(z - z0) = 0, where (x0, y0, z0) is the given point and (a, b, c) is the normal vector.

4. What is a normal vector?

A normal vector is a vector that is perpendicular to a line or plane. It is used to define the direction of the line or plane and is an essential component in the point-normal equation.

5. Are there any applications of point-normal equations?

Yes, point-normal equations have various practical applications in fields such as physics, engineering, and computer graphics. They are used to represent and calculate the properties of lines and planes in three-dimensional space, making them useful in solving real-world problems.

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