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A few weeks ago at school we started a course in linear algebra. A few days ago we got a problem and some steps on how to solve the problem. I get this working just plugging in numbers, but I don't understand why this actually works. The problem is described below:

find the intersection x y coordinates of the following equations:

$$ \begin{bmatrix}x\\y \\ \end{bmatrix} = \begin{bmatrix}6\\-\frac{8}{3} \\ \end{bmatrix} + \lambda\begin{bmatrix}12\\-6 \\ \end{bmatrix} \\ \begin{bmatrix}11\\ 2\\ \end{bmatrix}\begin{bmatrix}x\\y \\ \end{bmatrix}=\begin{bmatrix}11\\2 \\ \end{bmatrix}\begin{bmatrix}5\\-\frac{47}{2}\\ \end{bmatrix} $$

We were taught to set up the equation of a perpendicular line of the first equation, then calculating some constant $c$ : $$ -6x-12y=c \\ -6*6-12*-\frac{8}{3} = c \\ c=-4 $$

Then for the second equation, we get the following: $$ 11x+2y=11*5+2*-\frac{47}{2}\\ 11x+2y=8 $$ Next, we solve the following system: $$ \left\{ \begin{array}{c} -6x-12y=-4 \\ 12x+2y = 8 \end{array} \right. $$

Which gives the answer $x = \frac{11}{15}$ and $y = \frac{-1}{30}$

What I don't understand:

  • Why do we need to calculate perpendicular line equation to find the point where these two lines intersect? Why can't we just use or modify the original equation?
  • What does that constant $c$ represent and why do we need it?

I tried plugging in things in online graphing calculators but I can't get a clear visual representation of what is actually going on here.

Unfortunately, the quality of our teacher isn't very great and he tells us to just do the trick without understanding why we need to do these steps. Hope someone can clarify this for me and my class mates! Thank you in advance.

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  • $\begingroup$ Do you mean for those products in the second equation to be dot products? $\endgroup$ – amd Jun 27 '18 at 18:59
  • $\begingroup$ @amd yes! As far as I understand, when there's no symbol between two vectors next to eachother, it's equal to the dot product $\endgroup$ – Markinson Jun 27 '18 at 19:38
  • $\begingroup$ That is not a widely-used convention, although it does appear in certain fields. $\endgroup$ – amd Jun 27 '18 at 19:39
  • $\begingroup$ @amd that's good to know! Can you give an example of when it's different? $\endgroup$ – Markinson Jun 27 '18 at 19:40
  • $\begingroup$ Well, there’s every answer that you’ve gotten so far. $\endgroup$ – amd Jun 27 '18 at 19:42
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You have here two of the fundamental ways to represent a line in $\mathbb R^2$. The first is described by a parametric representation that uses a point $\mathbf p_0$ on the line and a direction vector $\mathbf v$ parallel to the line. Every point on the line can then be expressed in the form $\mathbf p_0+\lambda\mathbf v$, $\lambda\in\mathbb R$.

The second line is described implicitly by an equation in point-normal form. There are various ways to interpret it geometrically, but I find this one easiest to visualize: The perpendicular line through the origin to a line $\mathscr l$ is completely characterized by a direction vector $\mathbf n$. Per the preceding paragraph, a parameterization of that perpendicular line is simply $\lambda\mathbf n$. If $\mathscr l$ is translated so that it passes through the origin, the translated line $\mathscr l'$ has the property that the position vector $\mathbf x'$ of every point on it is orthogonal to $\mathbf n$. That is, $\mathscr l'$ is the set of points that satisfy $\mathbf n\cdot\mathbf x'=0$. We can translate $\mathscr l$ in this way by fixing some point $\mathbf p$ on it and subtracting this from every point $\mathbf x$ on $\mathscr l$. Substituting into the orthogonality condition, an equation for $\mathscr l$ is therefore $\mathbf n\cdot(\mathbf x-\mathbf p)=0$, or $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf p$. Observe that it doesn’t matter which point was chosen for $\mathbf p$—the last form of the point-normal equation says that the dot product of any point on $\mathscr l$ with the normal vector $\mathbf n$ is constant.

Multiplying both sides of that last equation by $\mathbf n/\|\mathbf n\|^2$, we have $${\mathbf n \cdot \mathbf x\over \mathbf n\cdot\mathbf n}\mathbf n = {\mathbf n \cdot \mathbf p\over \mathbf n\cdot\mathbf n}\mathbf n,$$ which says that every point of $\mathscr l$ has the same orthogonal projection onto the normal vector $\mathbf n$. The length of this projection is $|\mathbf n\cdot\mathbf p|/\|\mathbf n\|$, which is also the distance of the line from the origin. Thus, the constant term in the point-normal equation of a line is the distance to the origin multiplied by the length of the normal vector. The sign of the constant term tells you on which side of the origin relative to the direction $\mathbf n$ the line lies.

To convert from the point-direction form to point-normal form, you have to find a normal to the line, which is any vector perpendicular to the line’s direction vector, and vice-versa when converting from point-normal to point-direction. The constant term for the point-normal form is then found by computing the dot product of this normal vector with any point on the line, as described above.

It’s not really necessary to convert representations to solve this problem, though. The intersection of the two lines is a point on the first line that satisfies the equation of the second, so you could simply plug the generic expressions for $x$ and $y$ from the first into the second and solve for $\lambda$ without doing any conversion. I expect that your professor has you do the conversion because he wants you to practice solving systems of linear equations, which is a major motivation for linear algebra in the first place.

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If your line $L$ is parametrized by $\mathbf{x}(\lambda) = (x,y) = \mathbf{x}_0 + \lambda \mathbf{v}$ then the shortest path from $\mathbf{0} = (0,0)$ to a point on $L$ is a line from $\mathbf{0}$ in the direction of $\mathbf{v}'$, a vector perpendicular to $\mathbf{v}$. From the properties of dot product we know that \begin{equation} \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos \theta \end{equation} where $\theta$ is the angle between vector $\mathbf{a}$ and $\mathbf{b}$. Therefore for any point $\mathbf{x}$ on $L$ we have \begin{equation} \mathbf{x}\cdot \mathbf{v}' = |\mathbf{v}'||\mathbf{x}|\cos\theta = |\mathbf{v}|'D \end{equation} where $D = |\mathbf{x}|\cos\theta$ is the shortest distance from $\mathbf{0}$ to $L$. The most important part is that $\mathbf{x}\cdot \mathbf{v}'$ is constant for all $\mathbf{x}\in L$ so we can compute this constant by calculating $\mathbf{x}_0\cdot \mathbf{v}'$. So the equation for your line $L$ in terms of $x,y$ is given by \begin{equation} \mathbf{x}\cdot \mathbf{v}' = \mathbf{x}_0\cdot \mathbf{v}'. \end{equation} If you have two lines, you can find line equations for both of the lines this way then solve the simutaneous equation to find their intersection point.

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The general equation of a line is $Ax+By+C=0$ or $Ax+By=C$, since we do not care whether $C$ is positive or negative. We can write the equation in vector form as $$\left(\begin{array}{c}x\\y\end{array}\right)\cdot\left(\begin{array}{c}A\\B\end{array}\right)=C$$ The vector (A,B) formed from the coefficients will always be orthogonal to the line. (Aside: holds for 3d and the plane too. A vector formed from the coefficients of $Ax+By+Cz=D$ will be orthogonal to the plane.)

So if we have the direction vector of a line, we quickly obtain the normal vector and then solve for C to get the line. There is actually an even quicker way. One form of a line is called the "normal equation of the line". It is written as $$\left(\begin{array}{c}x\\y\end{array}\right)\cdot\left(\begin{array}{c}n_{x}\\n_{y}\end{array}\right)=\left(\begin{array}{c}P_{x}\\P_{y}\end{array}\right)\cdot\left(\begin{array}{c}n_{x}\\n_{y}\end{array}\right)$$This gets a Cartesian equation and avoids having to explicitely calculate C.

In your case, $$\left(\begin{array}{c}x\\y\end{array}\right)\cdot\left(\begin{array}{c}6\\12\end{array}\right)=\left(\begin{array}{c}6\\-\frac{8}{3}\end{array}\right)\cdot\left(\begin{array}{c}6\\12\end{array}\right)$$ $$6x+12y=36-\frac{96}{3}$$

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