The intersection of $w = x / \beta$ and a line parallel to $w = \beta x$.

Question

If you are familiar with physics, you may have guessed what the background is… but, the following is simply a question of elementary geometry. (Not consider the special relativity.)

Question:

Consider the $x$-$w$ plane and $w=x/\beta$ (Red line of Fig.1) and $w=\beta x$ (Green line of Fig.1)
Under the following setting,

• Find the coordinates of $B$ and,
• Find the length of $OB$ using $\lvert OA\rvert$ and $\beta$.

Detailed settings are as shown in the Figure.1 :

• $O$ is the origin of $x$-$w$ plane,
• $0<\beta<1$
• $A$ is a point on the $x$-axis,
• $\lvert OA\rvert$ represents the length of $OA$, and
• The purple line on Fig.1 is a line that passing through the point $A$ and parallel to $w = \beta x$
• $B$ is the intersection the purple line and $w = \beta x$.

Fig.1

I tried to find the coordinates of $B$ and the length of $OB$ as follows. Is it correct?

I tried to find…:
The vertical leg from $B$ to the $x$-axis be denoted by $h$.
We define k as follows: $$k=1/\beta$$ Considering the trigonometric ratio, we get following: $$\frac{\left(\left\lvert OA\right\rvert+\left\lvert Ah\right\rvert\right)}{\left\lvert Bh\right\rvert}=k \\ \frac{\left\lvert Bh\right\rvert}{\left\lvert Ah\right\rvert}=k.$$

Transform above equations step by step: $$\left(\left\lvert OA\right\rvert+\left\lvert Ah\right\rvert\right)=k\left\lvert Bh\right\rvert \\ \left\lvert Bh\right\rvert=k\left\lvert Ah\right\rvert \\ \left\lvert Ah\right\rvert=\frac{\left\lvert Bh\right\rvert}{k} \\ \left(k\left\lvert OA\right\rvert+\left\lvert Bh\right\rvert\right)=k^2\left\lvert Bh\right\rvert \\ k\left\lvert OA\right\rvert = \left(k^2-1\right)\left\lvert Bh\right\rvert.$$

We get, $$\left\lvert Bh\right\rvert = \left(\frac{k}{k^2-1}\right)\left\lvert OA\right\rvert \\ \left\lvert Ah\right\rvert = \left(\frac{1}{k^2-1}\right) \left\lvert OA\right\rvert,$$ and

$$\left\lvert Oh\right\rvert =\left(\left\lvert OA\right\rvert+\left\lvert Ah\right\rvert\right) =k\left\lvert Bh\right\rvert =\left(\frac{k^2}{k^{2}-1}\right) \left\lvert OA\right\rvert.$$ Therefore, the Cartesian coordinate of $B$ is: $$B =\left(\left(\frac{k^2}{k^{2}-1}\right) \left\lvert OA\right\rvert , \left(\frac{1}{k^2-1}\right) \left\lvert OA\right\rvert \right) =\left(\frac{\left\lvert OA\right\rvert}{\left(\frac{1-\beta^2}{\beta^2}\right)}\right) \left(\frac{1}{\beta^2}, 1\right) \\ = \left(\frac{\left\lvert OA\right\rvert}{1-\beta^2}\right)\left(1, \beta^{2}\right).$$

And using Pythagorean theorem, the length of $OB$ is calculated as follows. $$\left\lvert OB\right\rvert =\frac{\left\lvert OA\right\rvert\sqrt{1+{\beta}^{2}}}{1-\beta^2}.$$

Attachment: You can use PPT　version of Fig.1 from this link.

P.S. I'm not very good at English, so I'm sorry if I have some impolite or unclear expressions. I welcome any corrections and English review. (You can edit my question and description to improve them)

Answer 1

Your geometrically-derived result looks correct. I would’ve computed the intersection of the two lines directly instead.

The equation of the green line is, as you wrote, $w=\beta x$, and setting $A=(a,0)$, the equation of the purple line is $w=(x-a)/\beta$. Expressing these two equations as homogeneous coordinate vectors, we find their intersection by computing their cross product $$\left(\beta,-1,0\right)\times\left(\frac1\beta,-1,-\frac a\beta\right) = \left(\frac a\beta,a,\frac1\beta-\beta\right)$$ and then dehomogenizing (dividing through by the last coordinate) to obtain $$B = \left(\frac a{1-\beta^2},\frac{a\beta}{1-\beta^2}\right) = \left(\frac1{1-\beta^2},\frac\beta{1-\beta^2}\right) \lvert OA\rvert.$$ Therefore $$\lvert OB\rvert = {\sqrt{1+\beta^2} \over 1-\beta^2} \lvert OA\rvert,$$ which agrees with your solution.

Answer 2

Starting from the notations provided by @amd, the green line is $w=\beta x$, the red line is $w=x/\beta$, and the purple line is $w=(x-a)/\beta$. For the intersection point $B$, we use the fact that $x$ and $w$ verify equations from both green and purple lines simultaneously $$\begin{align}w&=\beta x\\w&=\frac{x-a}\beta\end{align}$$ This yields $$x-a=\beta^2x$$ which you can simplify to obtain $$x_B=\frac a{1-\beta^2}$$ for thew $w$ then $$w_B=\beta x_B$$ Then $$|OB|=\sqrt{x_B^2+w_B^2}=x_B\sqrt{1+\beta^2}=|OA|\frac{\sqrt{1+\beta^2}}{1-\beta^2}$$