Let $V = \text{span}{(1, 1, 1)} \subset R^3$. Find a basis for $V^\perp$
Let $v \in V, u \in V^\perp$ such that that $v\cdot u = 0$.
Hence $c(1, 1, 1)\cdot (u_1, u_2, u_3) = 0$ for some $c \in \mathbb{R}$. Therefore $ u_1 + u_2 + u_3 = 0$.
This is where I get stuck. I have two questions;
Is it correct that $V^\perp$ represents a plane perpendicular to V and passing through (0, 0, 0)?
Is my earlier working correct? How can I turn it into a basis for $V^\perp$?
Yes, $u_1 + u_2 + u_3 = 0 $ is a plane in $R^3$ The standard equation of a plane is $ Ax + By + Cz = D$ or $Ax + By +Cz + D = 0 $ (opposite signs on D depending on your preferred formulation). With your $u_1, u_2, u_3$ equivalent to $x, y, z$, clearly you have a plane.
Note you could save yourself trouble by knowing the fact that the normal to a plane $Ax + By + Cz = D$ is the vector $(A, B, C)$
Since your D = 0 yes your plane passes through the origin. D must be zero in order for the plane to be a subspace. (You can check this. If D is not zero closure under addition fails.)
To get basis vectors for this plane find two independent vectors which are orthogonal to (1, 1, 1) You can do this by simply choosing two out of the three coordinates differently for each vector and letting the third be zero. Note that in two dimensions $(b, -a)$ is orthogonal or perpendicular to $(a, b)$
Let $v_1 = (1, -1, 0) $ and let $v_2 = (0, 1, -1)$
First we see $v_1 \cdot (1, 1, 1) = 0$ and $v_2 \cdot (1, 1, 1) = 0$ so they are in the orthogonal space.
Then we test for independence
$ \pmatrix {1 & -1 & 0 \\ 0 & 1 & -1}$
Reduce. Add Row 1 + Row 2
$ \pmatrix {1 & 0 & -1 \\ 0 & 1 & -1}$
There are two pivots (non-zero leading elements which are the only non-zero entries in their columns) so dimension = 2 and the vectors are definitely independent.
This completes the answer to the question.
The plane $x + y + z = 0$ is the orthogonal space and
$v_1 = (1, -1, 0) $ , $v_2 = (0, 1, -1)$ form a basis for it.
Often we know two vectors and want to find the plane the generate. We use the cross-product $v_1 \times v_2$ to get the normal, and then the rule above to form the plane. It is worth working through this process with the above vectors to back-check and see how this all works.
Your answer to 1. is correct: you're looking for a plane perpendicular to $V$ (or equivalently, perpendicular to $v$). Your work for 2 is also correct. In particular, $V^\perp$ is the space solutions to the system of equations $$ \pmatrix{1&1&1}\pmatrix{u_1\\u_2\\ u_3} = 0 $$ That is, $V^\perp$ is the nullspace (kernel) of the matrix $\pmatrix{1&1&1}$. I would expect that you know how to compute such a basis at this point in the course.
If you require an orthonormal basis: take your solution and apply Gram-Schmidt.
Yes, you're correct.
For finding the basis of the space described by the equation $u_1 + u_2 + u_3 = 0$: you know that it's $2$-dimensional subspace, so you'll need two linearly independent vectors; if the fact that these can be chosen as $(1, 0, -1)$ and $(0,1,-1)$ is not obvious than think of the space of solutions to the equation $u_1 + u_2 + u_3 = 0$ - you take parameters $u, v \in \mathbb{R}$ and find that vectors of the form $(u, v,-u-v)$ are solutions and setting $(u,v) = (0,1)$ or $(u,v) = (1,0)$ yields the result.