Hello, can someone please tell me how the vectors V1 and V2 are computed for when lambda=2?
Say,
P' =
\begin{bmatrix}
(2-\lambda) && 0 && 0 \\
1 && (2 - \lambda) && 1\\
-1 && 0 && (1-\lambda)
\end{bmatrix}
Finally, when \lambda = 2,
P'_{\lambda=2} =
\begin{bmatrix}
0 && 0 && 0 \\
1 && 0 && 1\\
-1 && 0 && -1
\end{bmatrix}
However, we must have,
P'_{\lambda=2}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix} =
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
\implies \begin{bmatrix} 0 && 0 && 0 \\ 1 && 0 && 1\\ -1 && 0 && -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
\implies x \begin{bmatrix} 0 \\1\\ -1 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} + z \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
Now, how can we make both sides equal?
- Put x=z=0, and any y\in \mathbb{R}…we take the unit case…
- Put y=0, and (x=-z) \wedge x, z \in \mathbb{R} …again we take the unit case…
In this, way, we obtain \vec{V_1} from 1, and \vec{V_2} from 2.
Hope it helps!
Happy learning…^^
@AeryB Sorry but there are a lot of segments in your response that show Math Processing Error, and I couldn’t fully understand your explanation because of that.
I don’t know why it’s not rendered for you but hold on!
I’m sending you the screenshot!
Inform me if your doubt remains.
1 Like
Thanks @AeryB - it makes perfect sense with your images. Also, for some strange reason your earlier response was not rendering the images last evening, but it is now…which sounds like it might be a browser issue or an issue with whatever CDN is on the server side.
1 Like
@AeryB if you interested, this is how your post shows occasionally, and I have no idea why it does this or why is it only occasionally:
Thanks for the feedback.
Best wishes…^^