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Use the convention that the Hermitian inner product is conjugate-linear in its first argument and linear in its second. In orthonormal coordinates, $⟨v,w⟩={\sum }_i\overline{v_i}{w}_i$. It satisfies conjugate symmetry, sesquilinearity and positive definiteness. Expanding in the orthonormal basis givesApplying the Hermitian inner product and $⟨e_i,e_j⟩={\delta }_{ij}$ yieldsThis identifies exactly where complex conjugation enters; replacing the first factor by $⟨e_i,v⟩$ would generally be wrong under this convention.

Use the point ${\mathbf{p}}_0=(-2,-2,4)$ and the orthonormal basis of directions in the cutting planeBoth vectors are orthogonal to the normal vector $(1,1,2)$, so every point of the plane has unique coordinatesSubstituting into the cone equation givesConsequently the intersection has equationBecause these are orthonormal coordinates, their lengths are the actual lengths in the plane. Thus this is an ellipse with semi-major axis $\sqrt{24}$ and semi-minor axis $\sqrt{8}$, not merely a projected ellipse. By a, its eccentricity is

First use the preceding isometry $t$ to align the first two points, and put $R_3=t(P_3)$. It remains to align $R_3$ with $Q_3$ while fixing $Q_1,Q_2$. Use orthonormal coordinates in which $Q_1=(0,0)$ and $Q_2=(d,0)$, $d>0$, and write $R_3=(r,s)$, $Q_3=(u,v)$. The prescribed distances giveSubtracting yields $r=u$, and then $s^2=v^2$. Therefore $Q_3$ is either $R_3$ or its reflection across the line through $Q_1,Q_2$. In the first case take $s=t$; in the second compose $t$ with that reflection. Both constructions fix the first two images and send the third point to its required image. The argument also covers three collinear points, when the two possible third images coincide. Thus all three points can be matched by one plane isometry.

First use the preceding isometry $t$ to align the first two points, and put $R_3=t(P_3)$. It remains to align $R_3$ with $Q_3$ while fixing $Q_1,Q_2$. Use orthonormal coordinates in which $Q_1=(0,0)$ and $Q_2=(d,0)$, $d>0$, and write $R_3=(r,s)$, $Q_3=(u,v)$. The prescribed distances giveSubtracting yields $r=u$, and then $s^2=v^2$. Therefore $Q_3$ is either $R_3$ or its reflection across the line through $Q_1,Q_2$. In the first case take $s=t$; in the second compose $t$ with that reflection. Both constructions fix the first two images and send the third point to its required image. The argument also covers three collinear points, when the two possible third images coincide. Thus all three points can be matched by one plane isometry.

A unitary operator $U:H\to K$ is a surjective linear isometry between complex inner product spaces. It preserves every inner product, not just the norm of a selected vector:Its adjoint operator is its inverse function, soFor an operator on one space, both identities have the same identity operator on the right. In finite-dimensional orthonormal bases, its matrix is a unitary matrix; the real counterpart is an orthogonal transformation.

Surjectivity matters in infinite dimension. An isometric embedding can satisfy $U^{\ast }U=I$ without satisfying $U{U}^{\ast }=I$: a proper image leaves some vectors outside its range. A unitary operator instead gives a reversible change of orthonormal coordinates, preserving all lengths and orthogonality.