Let $$g=\begin{pmatrix}2\\-5\\-3\\-3\end{pmatrix}+\mathbb R\begin{pmatrix}1\\2\\3\\4\end{pmatrix}$$ and $$h=\begin{pmatrix}1\\-3\\0\\-1\end{pmatrix}+\mathbb R\begin{pmatrix}2\\3\\4\\5\end{pmatrix}.$$
- Find all lines that are perpendicular to both $g$ and $h$.
- Find the smallest affine subspace in $\mathbb R^4$ that contains both $g$ and $h$.
As for 1: One can easily see that the two lines are skew. Now, if $v_g$ and $v_h$ are the direction vectors of the lines I am first interested in a base of $U^\perp$ where $U=\langle v_g,v_h\rangle$. I got $$U^\perp=\left\langle\begin{pmatrix}2\\-3\\0\\1\end{pmatrix},\begin{pmatrix}-1\\1\\1\\-1\end{pmatrix}\right\rangle=:\langle v_1,v_2\rangle.$$ So now we should get two perpendecular lines $$l_1=p_1+\mathbb R v_1\quad\text{ and }\quad l_2=p_2+\mathbb R v_2$$ and need to find $p_1$ and $p_2$.
We can parametrize $g$ via $$ \vec{P}_{\lambda}=\left(\begin{array}{c} 2+\lambda\\ -5+2\lambda\\ -3+3\lambda\\ -3+4\lambda \end{array}\right) $$ and $h$ via $$ \vec{G}_{\mu}=\left(\begin{array}{c} 1+2\mu\\ -3+3\mu\\ 4\mu\\ -1+5\mu \end{array}\right). $$ So the connection of $g$ and $h$ has the direction vector $$ v=\overrightarrow{P_{\lambda}G_{\mu}}=\left(\begin{array}{c} -1+2\mu-\lambda\\ 2+3\mu-2\lambda\\ 3+4\mu-3\lambda\\ 2+5\mu-4\lambda \end{array}\right). $$ The condition $v\perp g$ and $v\perp h$ yields $$ \left\langle \left(\begin{array}{c} -1+2\mu-\lambda\\ 2+3\mu-2\lambda\\ 3+4\mu-3\lambda\\ 2+5\mu-4\lambda \end{array}\right),\left(\begin{array}{c} 1\\ 2\\ 3\\ 4 \end{array}\right)\right\rangle =0=\left\langle \left(\begin{array}{c} -1+2\mu-\lambda\\ 2+3\mu-2\lambda\\ 3+4\mu-3\lambda\\ 2+5\mu-4\lambda \end{array}\right),\left(\begin{array}{c} 2\\ 3\\ 4\\ 5 \end{array}\right)\right\rangle $$ and thus, $$ 20+40\mu-30\lambda=0\,\,\,\,\,\,\,\,\,\,\text{and}\,\,\,\,\,\,\,\,\,26+54\mu-40\lambda=0. $$ The solution of this system of linear equations is given by $\mu=1$ and $\lambda=2.$ With that, we find \begin{align*} l_{1} & =\vec{P}_{2}+\mathbb{R}\overrightarrow{P_{2}G_{1}}\\ & =\left(\begin{array}{c} 4\\ -1\\ 3\\ 5 \end{array}\right)+\mathbb{R}\left(\begin{array}{c} -1\\ 1\\ 1\\ -1 \end{array}\right). \end{align*} Is this correct so far? But how do I get the second one?
As for 2: For the smallest subspace that contains both $g$ and $h$ I would take $g+v$ where $v$ is the direction vector between $g$ and $h$ as mentioned above. Does this make sense?
Question 1
Your answer to the first question starting at We can parametrize $g$ via... looks good and you found the unique solution line.
The first part is wrong. You indeed computed well $U^\perp$. This space is of dimension $2$. That doesn't mean that there is two solutions. But just that the direction of the solutions belong to $U^\perp$.
Question 2
The smallest affine subspace is the one passing through $P_g$ and having for direction $Vect\{\vec{P_g P_h}, v_g,v_h\}$. This is an affine hyperplane.
The equation of such an affine hyperplane is
$$ax+by+cz+dt+e=0$$ and you need to find $a,b,c,d,e$. Which can be done by writing that $\vec{P_g P_h}, v_g,v_h$ belong to the associated vector hyperplane $ax+by+cz+dt=0$ while $P_g$ belongs to the affine hyperplane.
Leading (if I avoided computation mistakes...) to the equations $$\begin{cases} x &+2y &+3z &+4t & &= 0\\ 2x &+3y &+4z &+5t & &=0\\ -x &+2y &+3z &+2t & &=0\\ 2x &-5y &-3z &-3t &+ e &=0\\ \end{cases}$$
And finally to the affine hyperplane of equation $$-x +3y-3z+t+11=0$$