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I have been given the line equation of $y-8=0$, and have to find the unit normal vector? How would I go about doing this? is there a specific equation I need to use?

Following this I have been given the question: Find the equation of a line passing through point $A(9, −6)$ and orthogonal to vector $v=[−4,0]^T$. Which i am also unsure how to complete.

Any help would be much appreciated, it's been quite a while since i have done this sort of maths.

Thanks

robjohn
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Joe Perkins
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2 Answers2

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If the line is put in the form $$ ax+by=c\tag{1} $$ that is equivalent to $$ (a,b)\cdot(x,y)=c\tag{2} $$ which says that the points on the line are perpendicular to the vector $(a,b)$. Thus, the unit normals would be $$ \pm\frac{(a,b)}{|(a,b)|}\tag{3} $$ In the case of $y-8=0$, you get $0x+1y=8$, so the unit normals would be $$ \pm\frac{(0,1)}{|(0,1)|}=\pm(0,1)\tag{4} $$

Reversing the process in $(1)$ and $(2)$, we get that $$ -4x+0y=C\tag{5} $$ is perpendicular to $\begin{bmatrix}-4&0\end{bmatrix}^T$. Plug $\begin{bmatrix}9&-6\end{bmatrix}^T$ into $(5)$ to get $C=-36$ $$ \begin{align} -4x+0y&=-36\\ x&=9\tag{6} \end{align} $$ which would be the line orthogonal to $\begin{bmatrix}-4&0\end{bmatrix}^T$ and passing through $\begin{bmatrix}9&-6\end{bmatrix}^T$.

robjohn
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  • Is the relation b/w (a,b) and (x,y) just a clever observation here, or is it part of some (line) rule(s)? Can you point to more info, thanks. – user362515 Dec 05 '14 at 19:27
  • @user362515: $(1)$ and $(2)$ are just rewriting things. The fact that the points $(x,y)$ that satisfy $(2)$ lie in a line that is perpendicular to $(a,b)$ is a useful fact. This is actually extendible to higher dimensions. Suppose that $(x_0,y_0)$ and $(x_1,y_1)$ lie in a subspace that is perpendicular to (a,b). Then, $(a,b)\cdot(x_1-x_0,y_1-y_0)=0$. That is, $(a,b)\cdot(x_1,y_1)=(a,b)\cdot(x_0,y_0)=c$. – robjohn Dec 05 '14 at 21:52
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For the general case, to get a vector perpendicular to a given one, you want the dot product to be zero. That means a vector perpendicular to $[a,b]$ is $[-b,a]$. Then you have to scale it to make it a unit vector.

In your specific cases, there are zeros involved. Have you plotted $y-8=0$? You should be able to find a perpendicular vector by inspection.

For the second one, what is the direction of $v$? What directions are perpendicular to it?

Ross Millikan
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  • we arent given a direction of the vector, but i'd guess it would be perpendicular to the line i have to find, but im not sure how to find it as i have only been given one point it passes through. Thanks for your reply – Joe Perkins Apr 25 '13 at 22:41
  • @JoePerkins: The vector is $[-4,0]^T$ Presumably the root is the origin. Does that help? – Ross Millikan Apr 26 '13 at 02:47
  • so it would simply follow the x axis left 4 units? making the line that im trying to find y=0? – Joe Perkins Apr 26 '13 at 02:51
  • @JoePerkins: To the first question, yes. But then you need a vertical line, while $y=0$ is horizontal, and you need it to go through $(9,-6)$ – Ross Millikan Apr 26 '13 at 02:54
  • Ok I think i understand, My line is going to be from points (0,4), to (9,-6)? and from there i just need to find the equation which im pretty sure i can do. – Joe Perkins Apr 26 '13 at 08:24
  • The vectors orthogonal to $v$ are horizontal (and you lost a minus sign on $v$) but go through $(9,-6)$, so the line is $y=-6$. for the first, you line is $y=8$, which is horizontal, so the normal is vertical. A vertical unit vector is $(0,1)$ or $(0,-1)$ – Ross Millikan Apr 27 '13 at 04:55