Intersection of a hypersphere (4d) and a hyperplane I have a problem to solve and I really am in over my head here. Just to get me started, if I have a four dimensional plane and a 4 dimensional sphere
$ax + by + cz + dw = k$ and $x^2+y^2+z^2+w^2 = R$, and they intersect, am I expecting a three dimensional sphere. 
In particular, what if I simply have x = r as the hyperplane, where r is less than R?

Multiply the $4d$ sphere by $(a^2+b^2+c^2+d^2)$ & rearrange it to give
\begin{eqnarray*}
(ax+by+cz+dw)^2+(bx-ay+dz-cw)^2+(cx-dy-az+bw)^2+(dx+cy-bz-aw)^2=R^2(a^2+b^2+c^2+d^2)
\end{eqnarray*}
Now substitute the equation for the plane & let
\begin{eqnarray*}
X=bx-ay+dz-cw \\
Y=cx-dy-az+bw\\
Z=dx+cy-bz-aw
\end{eqnarray*}
and $\rho^2 =R^2(a^2+b^2+c^2+d^2)-k^2$
\begin{eqnarray*}
X^2+Y^2+Z^2= \rho^2
\end{eqnarray*}
So the manifold of intersetion is indeed a $3d$ sphere.
