Both b1 is in parallel with x1, therefore, it should be a constant coefficient such that x1=c1a. Suppose x1=[2, 4] and a=[-8, -16]. They are parallel so that constant c1 is -1/4.
@@utkarshkathuria2931 Good question:) I am using 2d plane for the sake of representation but what I am solving is the case in R^n. Every point no matter what the dimension is, either 2 or n, would be a vector. That is why I am saying point x1 is parallel with (a) since both are two vectors. In the mentioned sense we can talk about their angle. Please let met know if you have any other question.
@@mathelecs3884 oh I get it now. I can imagine it in 3-D and think of an 'a' vector passing parallel to point x, which appears to be passing through point x in 2-D. Is it right?
I am studying this book by myself. Your videos are of great help. Thanks!
Thank you for your comment! I am so grateful that you find my videos helpful.
thank you so much!! All these video is very useful. Looking forward to the solution for the remaining chapter.
good stuff, thank you for your time!
You are welcome!
At 6:40 you said there is a vector b?
How do you get x1=c1a?
Both b1 is in parallel with x1, therefore, it should be a constant coefficient such that x1=c1a. Suppose x1=[2, 4] and a=[-8, -16]. They are parallel so that constant c1 is -1/4.
@@mathelecs3884 But a is vector passing through point x1. How can a and x1 be parallel?
@@utkarshkathuria2931 Good question:) I am using 2d plane for the sake of representation but what I am solving is the case in R^n. Every point no matter what the dimension is, either 2 or n, would be a vector. That is why I am saying point x1 is parallel with (a) since both are two vectors. In the mentioned sense we can talk about their angle. Please let met know if you have any other question.
@@mathelecs3884 oh I get it now. I can imagine it in 3-D and think of an 'a' vector passing parallel to point x, which appears to be passing through point x in 2-D. Is it right?
@@utkarshkathuria2931 yes:) you got the point! and I am happy that I could help.