In $\mathbb{E^3}$ we have the plane $\pi:x-y+z-3=0$, the line $r:(2,0,1)+t(1,1,0),\ t\in\mathbb{R}$, and the point $P=(3,0,3)$. Which conditions must parameters $a$ and $b$ meet so there's exist an orthonormal basis $\bar R$ in which $\pi:\bar y=2,\ r:\{\bar x=0,\bar y=a\},\ P=(0,b,0)$? Which conditions must this basis $\bar R=\{q;u_1,u_2,u_3\}$ meet? How many orthonormal basis exist that verify all these conditons?
$\mathbf{Attempt}$
Let's search for the vectors $u_1,u_2,u_3$. Extracting information from the planes, we have that their normal vectors are $n_\pi=(1,-1,1)=(0,1,0)_\bar R$, so $u_2=\frac{1}{\sqrt 3}(1,-1,1)$, where $\frac{1}{\sqrt 3}=||(1,-1,1)||$.
We can write $r_\bar R$ as $r_\bar R:(0,a,0)+\lambda(0,0,1)$, and if we impose that $(1,1,0)=(0,0,1)_\bar R$, then we get $u_3=\frac{1}{\sqrt 2}(1,1,0)$. To obtain $u_1$, we can use cross product: $u_1=u_2\times u_3=\frac{1}{\sqrt 6}(-1,1,2).$
In order to find the value of the parameters $a$ and $b$ we can find the values that preserve the relative positions of $\pi,\ r,$ and $P$. $\quad r\subset \pi\to\bar y = a = 2,\quad P\notin\pi\to b\neq 2$
Now we find $q$. We know that $(3,0,3)$=$(0,b,0)_\bar R$. In $\bar R$ basis $(0,b,0)_\bar R=0+b(e_2)_\bar R$, so in the standard basis $(3,0,3)=q+\frac{b}{\sqrt 3}(1,-1,1)$, so $q=(3-\frac{b}{\sqrt 3},\frac{b}{\sqrt 3},3-\frac{b}{\sqrt 3})$.
Does this mean that there are infinite orthogonal basis as we can choose any random $b\in\mathbb{R}\setminus(2)$? Or am I making wrong assumptions? Thanks in advance!
Best Answer
If I understood your notation right, your solution is correct, up to minor remarks, but incomplete.
$\sqrt 3=||(1,-1,1)||$.
Also signs of $u_i$ can be arbitrary.
But there is a problem. We have
$$r=\{ (2,0,1)+t(1,1,0):t\in\mathbb{R}\}=\{ q+a u_2+\lambda u_3:\lambda\in\mathbb{R}\},$$
that is $q+au_2-(2,0,1)=xu_3$ for some $x\in\Bbb R$. Since $q=(3,0,3)-bu_2$, we have $(1,0,2)+(2-b)u_2+y(1,1,0)$ for some $y\in\Bbb R$. But this is impossible, because $\left|\begin{array}{} 1 & 0 & 2\\ 1 & -1 & 1\\ 1 & 1 & 0 \end{array} \right|=3\ne 0$, so the vectors $(1,0,2)$, $u_2$, and $u_3$ are linearly independent.
I tried to find an error in our arguments, but I don’t see it.