Linear Algebra show line touches plane in one point x We consider these vectors in $R^4$
$$u_1=\begin{pmatrix}
1 \\
1 \\
2 \\
1
\end{pmatrix},u_2=\begin{pmatrix}
1 \\
-1 \\
0 \\
-1
\end{pmatrix}$$
and
$$v_1=\begin{pmatrix}
1 \\
9 \\
10 \\
9
\end{pmatrix},u_2=\begin{pmatrix}
1 \\
7 \\
8 \\
7
\end{pmatrix}$$
And we have that $B=\{u_1,u_2\}$ and $C=\{v_1,v_2\}$
both are bases for the same subspace(plane) $U\subset R^4$
Now we
consider the set $L = \{t e_1 + e_3 | t ∈ R \}$, ie. line in $R^4$ through the point $e_3 = (0,0,1,0)$ with
direction vector $e_1 = (1,0,0,0)$.
Now I have to show that the line $L$
cuts the plane $U$ in exactly one point $x \in R^4$ and I have to find that point.
I find it a bit abstract to see that. Can someone help me? How can I show it and witch equation systems do I have to solve to find this point?
 A: You want to find $L \cap U$, by definition this set is exactly the points $x$ satisfying both :

*

*$x = e_3 + t e_1$ for some $t \in \mathbb{R}$

*$x = a u_1 + b u_2$ for some $a,b \in \mathbb{R}$
Now, if there is a solution, it has to satisfy $e_3 + t e_1 = a u_1 + b u_2$, so $e_3 = a u_1 + b u_2 - t e_1$. Let's solve this for $a,b,t$, which gives us the following equations (you can see this by writing out the latter equality's vectors down) :

*

*First coordinate : $0 = a + b - t$

*Second coordinate : $0 = a - b$

*Third coordinate : $1 = 2a$

*Fourth coordinate : $0 = a - b$
giving us $a=b= \frac{1}{2}$ and $t = 1$ necessarily if $x$ is a solution.
This yields that the only possible point $x \in U \cap L$ is $x = e_1 + e_3 = \frac{1}{2} (u_1 + u_2) = (1,0,1,0)$, and this indeed holds. So $x$ is the unique solution.
Notice when we solved our system of equations that $a,b,t$ got fixed to necessarily be specific values, yielding unicity of the solution. If there were more than one solution, we would have a free variable.
A: To prove the first claim, that $B$ and $C$ are bases for the same subspace, we can show that $B$ and $C$ are linearly independent (which is trivial), that both elements of $C$ are in the span of $B,$ and that each element of $B$ is in the span of $C.$
To make these easier, it helps to simplify things as much as possible. For example, let's consider the span of $B.$ Every element of the span of $B$ is of the form $$\vec v=a\vec u_1+b\vec u_2=\begin{pmatrix}
a+b \\
a-b \\
2a \\
a-b
\end{pmatrix}.$$ The clever observation that $2a=(a+b)+(a-b)$ lets us rewrite this as $$\vec v=(a+b)\begin{pmatrix}
1 \\
0 \\
1 \\
0
\end{pmatrix}+(a-b)\begin{pmatrix}
0 \\
1 \\
1 \\
1
\end{pmatrix},$$ so reparameterizing by $c=a+b$ and $d=a-b,$ we find that the elements of the span of $B$ are exactly those of the span of $D=\left\{\vec w_1,\vec w_2\right\},$ where $$\vec w_1=\begin{pmatrix}
1 \\
0 \\
1 \\
0
\end{pmatrix},\;\vec w_2=\begin{pmatrix}
0 \\
1 \\
1 \\
1
\end{pmatrix}.$$ I leave it to you to show that $\vec w_1$ and $\vec w_2$ are elements of the span of $C,$ and that the elements of $C$ are in the span of $D.$
Now, we take an arbitrary element $\vec v$ of $L\cap U.$ Since $\vec v$ is an element of $U,$ then it is in the span of $D,$ meaning that $$\vec v=r\vec w_1+s\vec w_2=\begin{pmatrix}
r \\
s \\
r+s \\
s
\end{pmatrix}$$ for some real $r,s.$ On the other hand, $\vec v$ is an element of $L$, so that $$\vec v=t\vec e_1+\vec e_3=\begin{pmatrix}
t \\
0 \\
1 \\
0
\end{pmatrix}.$$ Readily, we conclude that $s=0$ and $r=t=1,$ meaning $\vec v=\vec e_1+\vec e_3.$
