MATLAB Finding the Unit Vector in the Direction of a Vector using Matrices and MATLAB

  • Thread starter Thread starter andyintranzit
  • Start date Start date
  • Tags Tags
    Matlab Matrices
AI Thread Summary
The discussion centers on a physics test question regarding the calculation of a unit vector in the direction of vector v, starting from vector u, using MATLAB. The options provided include various mathematical expressions involving vector operations. The user initially expresses confusion about matrix algebra and eliminates options b and c, reasoning that option b involves multiplying two identical vector matrices, which is not valid, and questioning the existence of the mag() function in MATLAB. Another participant clarifies that mag() is not a MATLAB function and suggests that it may refer to abs(), which calculates the magnitude of a complex number. They confirm that option d is correct, explaining that the expression involves normalizing the vector (v-u) using the appropriate mathematical operations, including the dot product for row vectors.
andyintranzit
Messages
6
Reaction score
0
Hey guys,

Sorry if this is the wrong forum. Someone bump it across if its not.

This is a question on one of my tests (im doing my first semester of Physics at university) and its got to do with the program Matlab.


We have two vectors u and v. Which of the following gives us the unit vector in the direction of v, starting from u?

a) (v-u)/(u*u')
b) (v-u) * (v - u)
c) (v-u)./mag(v-u)
d) (v-u) / (sqrt(v-u)*(v-u)')




'I understand some basic matrix algebra, but I am still pretty spun out about matrices. can someone guide me through this? (a) thanks!
 
Physics news on Phys.org
ok.. i think i can eliminate the middle two because...

for b) you can't multiply two identical vector matrices together? cause they'll be in the same form?

and for c) I am pretty sure that mag() isn't a Matlab function at all...
 
The answer is d.

No mag() is not a MATLAB function, perhaps they meant abs() which is a function to return the magnetude of a complex value. In any case sqrt(v-u)*(v-u)') is the correct divisor to normalize it.

BTW you know that for row vectors x and y that x*y' gives the dot product of x and y, right? If you happen to be using column vectors then it's just the simple modification of using x'*y of course.
 
Last edited:

Similar threads

Replies
4
Views
3K
Replies
3
Views
4K
Replies
6
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Back
Top