Image of a Linear Transformation

[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)

 

[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 $T3 * T2$?

 

[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.)

 

[skiller]

The "-3" and "0" are correct

How is the "0" correct?

 

[micromass]

How is the "0" correct?

Yes, only the $-3$ is correct.

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

 

[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.