7 projection on a different axes question

  • Thread starter Thread starter nhrock3
  • Start date Start date
  • Tags Tags
    Axes Projection
Click For Summary
SUMMARY

The discussion focuses on the projection transformation T: R² → R², specifically projecting onto the X-axis parallel to the line y = -√3x. The transformation maintains the X-axis while altering the Y-axis based on the specified line. The representative matrices of T with respect to the basis B = {(1,0), (0,1)} can be derived by applying the transformation to the basis vectors, leading to T(1,0) = (1,0) and T(0,1) = k(1,√3), where k is determined through trigonometric analysis.

PREREQUISITES
  • Understanding of linear transformations in R²
  • Familiarity with matrix representation of transformations
  • Knowledge of trigonometric functions and their applications
  • Proficiency in vector operations and basis transformations
NEXT STEPS
  • Study the properties of linear transformations in R²
  • Learn how to derive transformation matrices from basis vectors
  • Explore trigonometric identities relevant to projection transformations
  • Investigate applications of projection transformations in computer graphics
USEFUL FOR

Mathematicians, physics students, computer graphics developers, and anyone interested in understanding linear transformations and projections in two-dimensional space.

nhrock3
Messages
403
Reaction score
0
7)
T:R^{2}->R^{2} projection transformation on X-axes parallel to the
line
y=-\sqrt{3}x
find the representative matrices of T{*} by B=\{(1,0),(0,1)\} basis
how i tried:
i understood that the x axes stayed the same but the y axes turned
into
y=-\sqrt{3}x
our T takes some vector and returns only the new x part with respect
to the new y axes.
any guidanse?
 
Physics news on Phys.org
nhrock3 said:
7)
T:R^{2}->R^{2} projection transformation on X-axes parallel to the
line
y=-\sqrt{3}x
find the representative matrices of T{*} by B=\{(1,0),(0,1)\} basis
how i tried:
i understood that the x axes stayed the same but the y axes turned
into
y=-\sqrt{3}x
our T takes some vector and returns only the new x part with respect
to the new y axes.
any guidanse?

If I understand what you're trying to say, then
T\begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}1\\ 0\end{bmatrix}
and
T\begin{bmatrix}0 \\ 1\end{bmatrix} = k\begin{bmatrix}1\\\sqrt{3}\end{bmatrix}

With a little trig you can figure out what k needs to be. What you know what a linear transformation does to a basis, you can write the matrix that represents the transformation.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K