Finding all vectors orthogonal to a vector I'm currently in a linear algebra course and I am just looking for feedback or some sort of verification of what I'm doing is correct.
So I have to find all vectors that are orthogonal to $u=(1, -2, 2, 1)$.
Seeing as this vector is in $R^4$, we let the vector $v=(v_1, v_2, v_3, v_4)$.
We also know that a vector is orthogonal to another, when the dot product of $u$ and $v$, $$u\cdot v=0$$
$$u\cdot v=(1, -2, 2, 1)\cdot(v_1, v_2, v_3, v_4)=v_1-2v_2+2v_3+v_4=0$$
This implies that $v_1=2v_2-2v_3-v_4$
Which means every vector that is orthogonal to the vector $(1,-2,2,1)$ will be in the form $v = (t, 2t, -2t, t)$ or $v = t(1,2,-2,1)$, letting $t$ be any real number.
 A: There is a problem in your conclusion. Everything is correct until you say that a vector $\overrightarrow{v}=(v_1,v_2,v_3,v_4)$ is orthogonal to the vector $\overrightarrow{u}=(1,-2,2,1)$ implies $v_1=2v_2-2v_3-v_4$.
From that point, the use of the $t$ is a bit weird: notice that the only thing we know is that given values for $v_2,v_3,v_4$, the value of $v_1$ will be completely determined. That is, the variable $v_1$ DEPENDS on the values of $v_2,v_3,v_4$, and the latter might be seen as INDEPENDENT variables.
If we write $v_2=r,v_3=s,v_4=t$ (which is simply a change of variables, and although popular it is completely unnecessary) we would get:
\begin{align*}
\overrightarrow{v}&=\begin{bmatrix}v_1\\ v_2\\ v_3\\ v_4\end{bmatrix}=\begin{bmatrix}2v_2-2v_3-v_4\\ v_2\\ v_3\\ v_4\end{bmatrix} &(\text{changing $v_1$ in terms of $v_2,v_3,v_4$)}\\
&=\begin{bmatrix}2r-2s-t\\r\\ s\\ t\end{bmatrix} &\text{(changing the variables)}\\
&=r\begin{bmatrix}2\\1\\0\\0\end{bmatrix} +s\begin{bmatrix}-2\\0\\1\\0\end{bmatrix}+t \begin{bmatrix}-1\\ 0\\ 0\\ 1\end{bmatrix}
\end{align*}
Hence, you can describe all the vectors that orthogonal to $\overrightarrow{u}=(1,-2,2,1)$ in several (equivalent) ways:


*

*Vectors of the form $\overrightarrow{v}=(2r-2s-t,r,s,t)$ where $r,s,t\in\mathbb{R}$.

*All linear combinations of the vectors $(2,1,0,0),(-2,0,1,0),(-1,0,0,1)$.

*Vectors in the subspace $W=\operatorname{span}\big((2,1,0,0),(-2,0,1,0),(-1,0,0,1)\big)$.

*Vectors of the form $(x,y,z,w)$ solving the equation $x-2y+2z+w=0$.

*Vectors contained in the hyperplane $x-2y+2z+w=0$.


Personally, I like the third and the fifth options.
A: The last conclusion is wrong,
$v_1 = 2v_2 - 2v_3 - v_4\space$implies that $V=\{(2a-2b-c,a,b,c);a,b,c\in \mathbb R \}$ 
