Interpolating points in a 3D space with one line

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Hello,
I have 5 points (black) in a 3D space. I would like to interpolate these points with a linear LINE and get the coefficients of this linear function.
I have just found how to interpolate these points with a surface so far by using the 'linearinterp' command in the curve fitting toolbox (see the attached figure).
Thanks for your help!

Accepted Answer

Star Strider
Star Strider on 4 Jul 2022
Try something like this —
rpm = rand(5,1);
dosage = rand(5,1)*0.1;
mass_of_leaves = rand(5,1);
B = [dosage mass_of_leaves ones(size(dosage))] \ rpm
B = 3×1
4.6860 0.3402 -0.0069
x = linspace(min(mass_of_leaves), max(mass_of_leaves), numel(mass_of_leaves));
y = linspace(min(dosage), max(dosage), numel(dosage));
z = [x(:) y(:) ones(size(x(:)))] * B;
figure
plot3(mass_of_leaves, dosage, rpm, 'p')
hold on
plot3(x, y, z, '-k')
hold off
grid on
xlabel('mass of leaves')
ylabel('dosage')
zlabel('rpm')
mdl = fitlm([mass_of_leaves dosage], rpm)
mdl =
Linear regression model: y ~ 1 + x1 + x2 Estimated Coefficients: Estimate SE tStat pValue _________ ________ ________ ________ (Intercept) -0.006867 0.061301 -0.11202 0.92104 x1 0.34018 0.086056 3.953 0.058441 x2 4.686 0.72903 6.4277 0.023359 Number of observations: 5, Error degrees of freedom: 2 Root Mean Squared Error: 0.0479 R-squared: 0.98, Adjusted R-Squared: 0.96 F-statistic vs. constant model: 49.2, p-value = 0.0199
B = mdl.Coefficients.Estimate
B = 3×1
-0.0069 0.3402 4.6860
[ypred,yci] = predict(mdl, [x(:) y(:)]);
figure
plot3(mass_of_leaves, dosage, rpm, 'p')
hold on
plot3(x, y, ypred, '-k')
plot3(x, y, yci, '--k')
hold off
grid on
xlabel('mass of leaves')
ylabel('dosage')
zlabel('rpm')
.

More Answers (2)

Jonas
Jonas on 4 Jul 2022
have a look into the polyfitn() function on FEX

Matt J
Matt J on 4 Jul 2022
Edited: Matt J on 4 Jul 2022
You can use linear3dFit() from this FEX submission
Also, a 3D line is not given by a single equation, so it is not clear what you mean by "the coefficients".

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