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

[LMHmedchem]

Hello,

I have some simple 3d data like,

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,

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.

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

 

[MarkFL]

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. :)

 

[LMHmedchem]

Thanks, that worked easily.

In excel, this looks like,

for the new x coordinate (x3)

=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