Problem 167
Hard Conics and Transformations

The point O' has coordinates (1,2,-1) in the system Oxyz .…

The point $O'$ has coordinates $(1,2,-1)$ in the system $Oxyz$. The basis vectors of the system $O'x'y'z'$ are $\vec a\{-2,0,1\}$, $\vec b\{-1,-1,3\}$, $\vec c\{-1,-1,1\}$. Express the old coordinates through the new ones and find the coordinates of $O$ in the new system.

Explanation of Method

Transformation formulas for a frame with a given origin and basis, and the coordinates of O in the new frame.

Step 1: Direct formulas

From $\overrightarrow{OM}=\overrightarrow{OO'}+x'\vec a+y'\vec b+z'\vec c$: \[x=1-2x'-y'-z',\qquad y=2-y'-z',\qquad z=-1+x'+3y'+z'\]

Step 2: Independence

\[ \det A=\begin{vmatrix}-2&-1&-1\\0&-1&-1\\1&3&1\end{vmatrix}=-4\ne0 \]

The basis is non-coplanar. Its orientation is opposite to that of $Oxyz$, since the determinant is negative.

Step 3: Inverse formulas

\[ A^{-1}=-\frac14\begin{pmatrix}2&-2&0\\-1&-1&-2\\1&5&2\end{pmatrix},\qquad \begin{pmatrix}x'\\y'\\z'\end{pmatrix}=A^{-1}\begin{pmatrix}x-1\\y-2\\z+1\end{pmatrix} \]

\[x'=\frac{-x+y-1}{2},\qquad y'=\frac{x+y+2z-1}{4},\qquad z'=\frac{-x-5y-2z+9}{4}\]

Step 4: Coordinates of $O$

Putting $x=y=z=0$: $O\left(-\tfrac12,\ -\tfrac14,\ \tfrac94\right)\ \text{in the second system}$.

Check with the direct formulas:

x y z $\vec a$ $\vec b$ $\vec c$ $O'(1,2,-1)$ $O$
The second system: origin $O'(1,2,-1)$ and basis vectors $\vec a,\vec b,\vec c$ (not unit, not orthogonal). The determinant $-4$ shows that they are non-coplanar, with orientation opposite to $Oxyz$.

Result

\[\boxed{\,x=1-2x'-y'-z',\qquad y=2-y'-z',\qquad z=-1+x'+3y'+z'\,}\]

\[\boxed{\,x'=\frac{-x+y-1}{2},\qquad y'=\frac{x+y+2z-1}{4},\qquad z'=\frac{-x-5y-2z+9}{4}\,}\]

\[\boxed{\,O\left(-\tfrac12,\ -\tfrac14,\ \tfrac94\right)\ \text{in the second system}\,}\]

Remark

The vectors are not unit vectors ($|\vec a|=\sqrt5$, $|\vec b|=\sqrt{11}$, $|\vec c|=\sqrt3$). “Unit vectors” in the statement should read “basis vectors”; the frame is affine.