How to find unit vector with a positive first component? Let $\mathbb R^4$ have the Euclidean inner product. Find a unit vector with a positive first component that is orthogonal to all three of the following vectors.
u $= (1,-1,3,0)$, v $= (6,1,0,1)$, w $= (1,0,7,1)$
I keep getting $(-14, -77,-21,11)$ but obviously that is not a vector with positive first component. Can anyone show me how this is done?
 A: The vector $(-14,-77,-21,11)$ is not orthogonal to $(1,-1,3,0)$ for example.
Take the following vector.  $$\frac{1}{\sqrt{2^2+11^2+3^2+23^2}}(2,11,3,-23).$$
A: If $(x,y,z,t)$ is orthogonal to $(1,-1,3,0)$, $(6,1,0,1),$ and $(1,0,7,1)$, then $$x-y+3z=0\tag1$$ $$6x+y+t=0\tag2$$ $$x+7z+t=0.\tag3$$
To eliminate $t$, subtract $(3)$ from $(2)$ to get $$5x+y-7z=0\tag4,$$ and  then, to eliminate $y$, add $(1)$ and $(4)$ to get $$6x-4z=0\tag5.$$
We will arbitrarily set $x=2$ to make the arithmetic easy and then scale later.
From $(5)$, we then get $z=3$; from $(1)$, we then get $y=11$,
and, from $(2)$ or $(3)$, we then get $t=-23$.
So we have $(x,y,z,t)=(2,11,3,-23)$, and now it just needs to be normalized.
I think this is essentially what Michael Rozenberg did, but with more explanation.
A: Just solve the linear system system
$$\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \\ \mathbf{w}\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ t\end{bmatrix} = \begin{bmatrix} 1 & -1 & 3 & 0 \\ 6 & 1 & 0 & 1 \\ 1 & 0 & 7 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ t\end{bmatrix}$$
to obtain $$(x,y,z,t) = \alpha (2, 11, 3, -23), \quad\text{ for any $\alpha
\in\Bbb{R}$}.$$
Now pick any $\alpha > 0$ and there is your vector $(x,y,z,t) \perp \mathbf{u}, \mathbf{v}, \mathbf{w}$ with $x>0$.
