# Finding point-normal equations

• Computnik
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.

## 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?

-

See above.

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?

-

## 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.

1 person
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!

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.

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?

Last edited:
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.

1 person
I'm a bit lost. :/ Any more tips?

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...

You say you are taking linear algebra but the problems you are having are all much more basic algebra matters.

$$\frac{x-a}{d_1}= \frac{y- b}{d_2}$$

An obvious thing to do is to multiply both sides by $d_1d_2$ to get rid of the fractions:
$$d_2(x- a)= d_1(y- b)$$
$$d_2(x- a)- d_1(y- b)= 0$$

HallsofIvy said:
You say you are taking linear algebra but the problems you are having are all much more basic algebra matters.

$$\frac{x-a}{d_1}= \frac{y- b}{d_2}$$

An obvious thing to do is to multiply both sides by $d_1d_2$ to get rid of the fractions:
$$d_2(x- a)= d_1(y- b)$$
$$d_2(x- a)- d_1(y- b)= 0$$
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.

## 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.