Image of a Linear Transformation

1. Mar 24, 2014

pondzo

T2 projects orthogonally onto the xz-plane

T3 rotates clockwise through an angle of 3π/4 radians about the x axis

The point (-3, -4, -3) is first mapped by T2 and then T3. what are the coordinates of the resulting point?

this question is on a program call Calmaeth. My answer for this question is (-3,0,-√2/2). The program says its wrong but i have checked thoroughly many times and cannot find my mistake.

My transformation matrix for T2 is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ and for T3 is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{-1}{√2} & \frac{1}{√2} \\ 0 & \frac{-1}{√2} & \frac{-1}{√2} \end{pmatrix}$
To get the resulting standard matrix, i did T3*T2 and then multiplied this matrix by the point (-3, -4, -3) to get the resulting point.

Can anyone see where i went wrong if i did? (also to let you know, the program said my matrices for T2&T3 were correct)

2. Mar 24, 2014

jbunniii

Your method is fine. Assuming your matrices are right, it looks like you made an error with the matrix multiplication. What did you get for $T3 * T2$?

3. Mar 24, 2014

HallsofIvy

I agree with jbuniii. The "-3" and "0" are correct but I do not get $-\sqrt{2}/2$ as the third component when I multiply your matrices.

(As a check, rather than multiplying T3*T2 first and then multiplying that by the vector, you can multiply the vector by T2 and then multiply the result by T3.)

Last edited by a moderator: Mar 24, 2014
4. Mar 24, 2014

skiller

How is the "0" correct?

5. Mar 24, 2014

micromass

Yes, only the $-3$ is correct.

Anyway, please post this in the homework forum next time! Thanks

6. Mar 25, 2014

pondzo

Hi guys It was a rookie mistake on my part. I was doing [-3,-4,-3]*[standard matrix] . when i should have been doing [standard matrix]*$\begin{pmatrix} -3 \\ -4 \\ -3 \end{pmatrix}$
And sorry, in the future ill post in the homework section.