vector-parametric representations

[thepillow]

After solving a system of linear equations I have the following vector-parametric representation of the 2-dimensional solution space:

\[F:\begin{bmatrix}
v\\
w\\
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
2\\
0\\
2\\
0\\
0
\end{bmatrix}+t_{1}\begin{bmatrix}
-1\\
1\\
-1\\
1\\
0
\end{bmatrix}+t_{2}\begin{bmatrix}
0\\
0\\
0\\
0\\
1
\end{bmatrix} : t_{1}, t_{2}\in \mathbb{R}\]

Then we're asked to find a 3-dimensional linear surface, call it H, that intersects our 2-dimensional surface, F, in exactly a line.

Below is my attempt, and I was hoping for some feedback on whether my answer and methodology are correct:

I think that a 3-dimensional linear surface, H, that intersects F in exactly a line can be represented as
\[H:\begin{bmatrix}
v\\
w\\
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
2\\
0\\
2\\
0\\
0
\end{bmatrix}+t_{1}\begin{bmatrix}
-1\\
1\\
-1\\
1\\
0
\end{bmatrix}+\lambda_{1}\overrightarrow{\mathbf{v_{1}}}+\lambda_{2}\overrightarrow{\mathbf{v_{2}}}\]
where the lambdas are parameters and the vectors v₁ and v₂ are two vectors not in F.

If this is correct then a possible H could be:
\[H:\begin{bmatrix}
v\\
w\\
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
2\\
0\\
2\\
0\\
0
\end{bmatrix}+t_{1}\begin{bmatrix}
-1\\
1\\
-1\\
1\\
0
\end{bmatrix}+\lambda_{1}\begin{bmatrix}
1\\
0\\
-1\\
0\\
0
\end{bmatrix}+\lambda_{2}\begin{bmatrix}
1\\
1\\
0\\
0\\
0
\end{bmatrix} : t_{1},\lambda_{1},\lambda_{2}\in \mathbb{R}\]
because the two vectors\[\begin{bmatrix}1\\
0\\
-1\\
0\\
0
\end{bmatrix}, \begin{bmatrix}
1\\
1\\
0\\
0\\
0
\end{bmatrix}\]
are normal to both of the vectors in F associated with parameters.

Would this H indeed represent a 3-dimensional linear surface that intersects F in exactly a line?

Thanks in advance for your help and insights!

 

[daon2]

Looks good to me! Just be sure to show your spanning vectors in F and v1, v2 are linearly independent

 

[thepillow]

Looks good to me! Just be sure to show your spanning vectors in F and v1, v2 are linearly independent

Thanks daon2! I think I'm starting to get the hang of this stuff.

Regarding proving linear independence, is it sufficient to show that my spanning vectors in F are each pairwise orthogonal to v1, v2 since sets of pairwise orthogonal (non-zero) vectors are linearly independent?

For example, if I call my spanning vectors in F b1 and b2 and then I show this:

\[\begin{bmatrix}
-1\\
1\\
-1\\
1\\
0
\end{bmatrix}\cdot\begin{bmatrix}1\\
0\\
-1\\
0\\
0
\end{bmatrix}=0\: \therefore \mathbf{b_{1}}\perp \mathbf{v_{1}}
\]

\[\begin{bmatrix}
-1\\
1\\
-1\\
1\\
0
\end{bmatrix}\cdot \begin{bmatrix}1\\
1\\
0\\
0\\
0
\end{bmatrix}=0\: \therefore \mathbf{b_{1}}\perp \mathbf{v_{2}}
\]

and then show the same thing for b2 with each of v1, v2, is that sufficient proof of linear independence?


A million thanks!

 

[daon2]

Yes, (non-zero) orthogonal vectors are linearly independent, so that will do the job. You could prove this fact without too much effort.