Exam 2025 Question 28

Exam 2025 Question 28

by Luis Falke -
Number of replies: 1

The assumption looks wrong to me. I think [R∣t] is 3×4, but X is only 3×1, so we can't compute [R∣t]X.

In my understanding, this is the exact reason why we use homogeneous coordinates, so we can have this efficient way of computing.

So it is extra confusing for me why you then change the formula to: ~x2=K(R~X + t), since this also does not work as R is 3x3 and ~X is 4x1, and on top of that, it is the inefficient way of computing where we apply translation separately from the multiplication. So what we actually want is ~x2 = K[R∣t]~X

Am I missing something here?

In reply to Luis Falke

Re: Exam 2025 Question 28

by Yingxuan You -
You are right. The original notation was not precise enough.

[R|t]X is valid only when X is written as a homogeneous 3D point, i.e., X_bar = [X; 1], which is a 4-dimensional vector. If X denotes the 3D point in R^3, then the correct form is RX + t.

So your dimensionality concern is correct. The original solution mixed the point X in R^3 and the homogeneous point X_bar in R^4. We have updated the notation in the answer file and re-uploaded the Exam 2025 Answer file.