[While this may be unfamiliar...] the $ct'$ and $x'$ are Minkowski-perpendicular to each other in all frames.
In the reference frame of the $ct'$-axis, the axes will also appear to be [ordinary-Euclidean]-perpendicular to each other.
Here's how Minkowski describes this...
From Minkowski's "Space and Time"...
We decompose any vector, such as that from O to x, y, z, t into four
components x, y, z, t. If the directions of two vectors are, respectively, that
of a radius vector OR from O to one of the surfaces ∓F = 1, and that of
a tangent RS at the point R on the same surface, the vectors are called
normal to each other. Accordingly,
$$c^2tt_1 − xx_1 − yy_1 − zz_1 = 0$$
is the condition for the vectors with components x, y, z, t and $x_1$, $y_1$, $z_1$, $t_1$ to
be normal to each other.
In other words,
locate the intersection of an observer's 4-velocity with the unit-hyperbola (the Minkowski circle) centered at the tail of the observer's 4-velocity.
The tangent line to that hyperbola is Minkowski-perpendicular to that observer's 4-velocity. That observer's x-axis is drawn through the tail of her 4-velocity, parallel to that tangent line.
The "intuition" to have is that
the tangent to the "circle" in that geometry
is
orthogonal to the radius vector.
You can play around with this idea in my visualization [screencaptured below].
- The unit-hyperbola (the "Minkowski circle") is in blue.
This figure is unchanged by a Lorentz boost.
- The red dotted line is the observer-worldline.
- The red tangent line is the prototype for "simultaneous events according to
the red-observer".
- The red-observer's x-axis is drawn parallel to that tangent line.
https://www.desmos.com/calculator/r4eij6f9vw
Play around with the E-slider to see the Galilean and Euclidean analogues!
