Why is the transpose of N necessary to get the point?

I think the transpose is just to make the product typecheck. That is, since points are stored as row vectors, one must be transposed into a column to enable taking a product.

Can someone explain the +tN portion of this?

@sushain. Correct y^Tx is the matrix product of the 1x3 vector y^T and 3x1 vector x, which is the dot product.

@azwarens: Remember, points on the line are all points x such that N^Tx = c.

The direction from P directly toward the line is the opposite of the normal: -N. Therefore the value of t such that x=P-tN produces a point that satisfies N^Tx = c gives the distance from the point to the line.

So does this mean that N^T(p+tN)=c should be N^T(p-tN)=c in the slides above?

@tap22sf. Either is fine. Note that with the equation on the slide, you'd just solve for a negative value of t. However, I agree that it would have been a little more clear had I gone with the P-tN formulation on the slide.

Wouldn't $$d = |(p - x_0)^TN|$$ be a cleaner solution? We just project the $p$ vector onto the normal and check its magnitude.

What exactly is the x vector in this equation representing?

@m11 I think the x vector represents points on the line represented as vectors. So any x such that NˆTx=c is on the line

didn't get email notification...@kayvonf thank you that makes sense!

Just to check my understanding of this thread, does t represent the distance between p and the line, and does p - tN equal the "closest point" on the line?