MHB Formula to extend a line in 3d (opposite of midpoint formula)

  • Thread starter Thread starter LMHmedchem
  • Start date Start date
  • Tags Tags
    3d Formula Line
LMHmedchem
Messages
20
Reaction score
0
Hello,

I have some simple 3d data like,

Code:
point 1: x1 = 5.8573375, y1 = -4.17885, z1 = 2.338175
point 2: x2 = -3.26132, y2 = 1.28276, z2 = 0.931583

I need to be able to extend the line from point 1 through point 2 to some additional distance.

There are a few posts on stackexchange, but they are all in c++ code and I couldn't get results in excel that look correct. Maybe it's just too late in the day. What I found was more or less,

Code:
x3 = x2 * M - x1, y3 = y2 * M - y1, z3 = z2 * M - z1 where M is a multiplier

For the example I gave above, this gives me a point that is no where near anything that resembles an extension of the line from point 1 to point 2.

Code:
x3 = 2.92866875, y3 = 4.82023, z3 = -1.8723835

Maybe I just got the parenthesis wrong or something lame like that. This seems like it should be very simple but I'm not sure if I need to find the slope first or if something like the above will work.

Can someone point me to the correct formula?

Thanks,

LMHmedchem
 
Mathematics news on Phys.org
If you have a line in 3 dimensions specified by the two points:

$$\left(x_1,y_1,z_1\right),\,\left(x_2,y_2,z_2\right)$$

Then a vector along the line can be given by:

$$\vec{v}=\left[\begin{array}{c}x_1+\left(x_2-x_1\right)t \\ y_1+\left(y_2-y_1\right)t \\ z_1+\left(z_2-z_1\right)t \end{array}\right]$$

You can then use the parameter $t$ to generate additional points on the line. :)
 
Thanks, that worked easily.

In excel, this looks like,

for the new x coordinate (x3)
Code:
=A2+(A4-A2)*$D$2

where,
A2 is the cell containing x1
A4 is the cell containing x2
$D$2 is cell containing the multiplier

If the multiplier is < 1.0, the new point will fall between point 1 and point 2. A multiplier value of 0.5 would give you the same result as using the midpoint formula. The multiplier must be > 1.0 for point 3 to be beyond point 2. I would guess that if the multiplier is negative, point 3 would occur on the line from point 1 to point 2 but before point 1 but I haven't confirmed that.

I have a similar question but I will put that in a new thread because it is likely a more complex solution.

Thanks again,

LMHmedchem
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top