6
$\begingroup$

By a rotation in ${\bf R}^3$ I mean an orthogonal linear transformation $f:{\bf R}^3\to {\bf R}^3$ represented by a matrix $A$ (i.e. $fx=Ax$) with $\det A=1$. By a reflection (through $S$) I mean an orthogonal linear transformation such that for some subspace $S$, $f\mid S={\rm id}$ and $f\mid S^{\perp}=-{\rm id}$. I am currently stuck in two tasks:

$(1)$ Let $\mathscr L_1=\langle(1,1,0)\rangle+(2,0,1)$ and $\mathscr L_2=\langle (2,1,3)\rangle+(1,0,4)$. I have to find a rotation such that $f(\mathscr L_1)=\mathscr L_2$. Now, the distance of both lines to the origin is $\sqrt 3$. The first line accomplishes this with $(1,-1,1)$, while the second line accomplishes this with $(-1,-1,1)$. I tried various times to define a rotation, but I failed. In particular, I know I should map $(1,-1,1)$ to $(-1,-1,1)$.

$(2)$ Let $\Pi_1=\{(x_1,x_2,x_3):x_1-x_2+2x_3=k\}$ and $\Pi_2=\langle (1,0,1),(0,1,2)\rangle+(1,-1,1)$. I have to find $k$ such that there exists a reflection that maps $\Pi_1$ to $\Pi_2$ and find $f(\Pi_2)$. Now, the distance to $\Pi_1$ to the origin is $|k|/\sqrt 6$ and that of $\Pi_2$ is $3/\sqrt 6$ which gives me $k=3,-3$. I don't really know how to continue now.

$\endgroup$
  • $\begingroup$ I am confused. By definition there should be a $\textbf{unique}$ point in the line with the minimal distance to the origin. How come you get 2 points? $\endgroup$ – Bombyx mori Dec 6 '13 at 2:40
  • $\begingroup$ @user32240 Sorry, fixed. $\endgroup$ – Pedro Tamaroff Dec 6 '13 at 2:46
  • $\begingroup$ I think you misunderstood my comment: I have always seen these definitions for rotation and reflection. These require the linear transformations to be isometric. $\endgroup$ – Julien Dec 6 '13 at 2:58
  • $\begingroup$ @julien Sorry, I missed to state that. $\endgroup$ – Pedro Tamaroff Dec 6 '13 at 3:01
  • $\begingroup$ Couldn't you try solving (the first problem to start with) with respect to the basis made of the two directional vectors plus one (orthogonal to the plane panned by the other two) $\endgroup$ – AnyAD Dec 6 '13 at 3:11
5
$\begingroup$

(1) You found the orthogonal projections of the origin onto these lines. So we have $$ \mathcal{L}_1=p_0+\mathbb{R}u \quad\mathcal{L}_2=q_0+\mathbb{R}v\quad \mbox{with}\;p_0=\pmatrix{1\\-1\\1}\;u=\pmatrix{1\\1\\0}\;q_0=\pmatrix{-1\\-1\\1}\;v=\pmatrix{2\\1\\3} $$ We are looking for a rotation, that is a direct (determinant one) isometry $f$ such that $f(\mathcal{L}_1)=\mathcal{L}_2$ or, equivalently, $f(p_0)=q_0$ and $f(u)=\lambda v$. For such a rotation to exist, it is necessary and sufficient that $\|p_0\|=\|q_0\|$, which is the case here.

Let us normalize these vectors $$p_1=\pmatrix{1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}}\;\qquad u_1=\pmatrix{1/\sqrt{2}\\1/\sqrt{2}\\0}\;\qquad q_1=\pmatrix{-1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}}\;\qquad v_1=\pmatrix{2/\sqrt{14}\\1/\sqrt{14}\\3/\sqrt{14}} $$ By construction, $\{p_1,u_1\}$ and $\{q_1,v_1\}$ are pairs of orthogonal unit vectors. So $$ \{p_1,u_1,p_1\times u_1\}\quad\mbox{and}\quad \{q_1,v_1,q_1\times v_1\}\quad \mbox{where}\;p_1\times u_1=\pmatrix{-1/\sqrt{6}\\1/\sqrt{6}\\2/\sqrt{6}} \; q_1\times v_1=\pmatrix{-4/\sqrt{42}\\5/\sqrt{42}\\1/\sqrt{42}} $$ are two direct orthonormal bases. Therefore $$ f(p_1):=q_1\qquad f(u_1):=v_1\qquad f(p_1\times u_1):=q_1\times v_1 $$ defines the (not unique) direct isometry we are looking for. After a bit of tedious computations, we find the following matrix form for $f$ with respect to the canonical basis $$ f=\pmatrix{-\frac{1}{3}+\frac{5}{3\sqrt{7}} & \frac{1}{3}+\frac{1}{3\sqrt{7}} & -\frac{1}{3}-\frac{4}{3\sqrt{7}}\\ -\frac{1}{3}+\frac{1}{3\sqrt{7}}& \frac{1}{3}+\frac{4}{3\sqrt{7}} & -\frac{1}{3}+\frac{5}{3\sqrt{7}} \\ \frac{1}{3}+\frac{4}{3\sqrt{7}}& -\frac{1}{3}+\frac{5}{3\sqrt{7}} & \frac{1}{3}+\frac{1}{3\sqrt{7}}} $$

(2) In dimension $3$, reflections are isometries with eigenvalues $\{1,1,-1\}$.

Denote $\Pi_1=p+u^\perp$ and $\Pi_2=q+v^\perp$ where $$ p=\pmatrix{k\\0\\0} \quad u=\pmatrix{1\\-1\\2}\quad\qquad q=\pmatrix{1\\-1\\1}\quad v=\pmatrix{-1\\-2\\1} $$ where $v$ is the cross product of the two vectors spanning the linear plane in the definition of $\Pi_2$. The orthogonal projections of the origin onto $\Pi_1$ and $\Pi_2$ are $$ p_0=\pmatrix{k/6\\-k/6\\k/3}\qquad q_0=\pmatrix{-1/3\\-2/3\\1/3} $$

Note that now $$ \Pi_1=p_0+ p_0^\perp \qquad\mbox{and}\qquad \Pi_2=q_0+q_0^\perp $$ For an isometry $f$ to take $\Pi_1$ onto $\Pi_2$, it is necessary and sufficient that $$ f(p_0)=q_0\qquad \mbox{and}\qquad f(p_0^\perp)=q_0^\perp $$ Now note that a reflection $f$ is an involution and has $f^*=f=f^{-1}$. It follows that for a reflection $f$ $$ f(\Pi_1)=\Pi_2\quad\iff\quad f(\Pi_2)=\Pi_1 \quad \iff \quad f(p_0)=q_0 $$

Moreover, such an $f$ leaves the line $p_0^\perp\cap q_0^\perp$ invariant. This means that if we consider the vector $z:=p_0\times q_0$, we have $f(z)=\pm z$. But the case $f(z)=-z$ is impossible here since this would force $f$ to be the identity on $z^\perp$, whence $p_0=q_0$. So we must have $f(z)=z$.

For $f$ to exist, we need $\|p_0\|=\|q_0\|$, which amounts to $k=\pm 2$ in our case. It is also sufficient.

In any case, let us normalize our vectors $$ p_1:=\frac{p_0}{\|p_0\|}\quad p_1:=\frac{p_0}{\|p_0\|}\quad z_1:=\frac{p_0\times q_0}{\|p_0\times q_0\|} $$ and consider the following direct orthonormal bases $$ \{p_1,z_1,p_1\times z_1\}\qquad \mbox{and}\qquad \{q_1,z_1,q_1\times z_1\} $$ We are looking for a reflection which fixes $z_1$, and sends $p_1$ onto $q_1$. Drawing a picture in the plane $z_1^\perp$ where $p_1,q_1,p_1\times z_1, q_1\times z_1$ all lie, we see that we have no choice: we must have $f(p_1+q_1)=p_1+q_1$ and $f(p_1-q_1)=q_1-p_1$ whence $f(p_1\times z_1)=-q_1\times z_1$. And indeed $$ f(p_1):=q_1\qquad f(z_1):=z_1\qquad f(p_1\times z_1):=-q_1\times z_1 $$ does define the reflection we are looking for. And it is unique.

  • For $k=-2$, we get $$ p_1=\pmatrix{-1/\sqrt{6}\\1/\sqrt{6}\\-2/\sqrt{6}}\;q_1=\pmatrix{-1/\sqrt{6}\\-2/\sqrt{6}\\1/\sqrt{6}} \; z_1=\pmatrix{-1/\sqrt{3}\\1/\sqrt{3}\\1/\sqrt{3}} \; p_1\times z_1=\pmatrix{1/\sqrt{2}\\1/\sqrt{2}\\0} \;q_1\times z_1=\pmatrix{-1/\sqrt{2}\\0\\-1/\sqrt{2}} $$ So our reflection is $f(x,y,z)=(x,z,y)$.

  • For $k=2$, we get $$ p_1=\pmatrix{1/\sqrt{6}\\-1/\sqrt{6}\\2/\sqrt{6}}\;q_1=\pmatrix{-1/\sqrt{6}\\-2/\sqrt{6}\\1/\sqrt{6}} \; z_1=\pmatrix{1/\sqrt{3}\\-1/\sqrt{3}\\-1/\sqrt{3}} \; p_1\times z_1=\pmatrix{1/\sqrt{2}\\1/\sqrt{2}\\0} \;q_1\times z_1=\pmatrix{1/\sqrt{2}\\0\\1/\sqrt{2}} $$ In the canonical basis, we find that the matrix of $f$ is

$$ f=\pmatrix{-1/3&-2/3&-2/3\\-2/3&2/3&-1/3\\-2/3&-1/3&2/3} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.