# Finding point-normal equations

1. Oct 9, 2013

### Computnik

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
-

3. The attempt at a solution
See above.

2. Oct 9, 2013

### LCKurtz

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.

3. Oct 9, 2013

### Computnik

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. Oct 9, 2013

### LCKurtz

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.

5. Oct 9, 2013

### Computnik

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: Oct 9, 2013
6. Oct 9, 2013

### LCKurtz

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.

7. Oct 9, 2013

### Computnik

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

8. Oct 10, 2013

### Computnik

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. Oct 10, 2013

### HallsofIvy

Staff Emeritus
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:
$$d_2(x- a)= d_1(y- b)$$
$$d_2(x- a)- d_1(y- b)= 0$$

10. Oct 10, 2013

### Computnik

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.