Find a plane perpendicular to the intersection of two affine subspaces

Question

Consider the two following affine subspaces of $\mathbb{R^3} $: $$S=\{x,y,z)\mid 2x+y+z=1\}$$ and $$T=\{(x,y,z)\mid x-y+2z=0\} $$ Find the plane $H$ perpendicular to the intersección of $S$ and $T$, that passes through the point $p(-1,0,1)$.

Answer 1

The intersection of the two given planes is a line. A plane perpendicular to this line, is perpendicular to both planes $S$ and $T$. This means the scalar product (aka dot product) of the normal (vector) of the plane $H$ you're looking for, with both of the normal vectors of planes $S$ and $T$, has to be zero.

Or, even easier if you know the vector product: the vector product of the normal vectors $\vec n_S$ of $S$ and $\vec n_T$ of $T$ produce a vector perpendicular to both, so precisely a suitable normal vector for the plane $H$: $$\vec n_H = \vec n_S \times \vec n_T$$

Once you have found a normal vector of $H$, you already have $a$, $b$ and $c$ in the standard cartesian equation for $H$ ($ax+by+cz+d=0$), use the given point $p$ to find $d$.

Or you may have seen a standard form for a cartesian equation of a plane with normal vector $\vec n = (a,b,c)$ and through the point $p(x_0,y_0,z_0)$ as: $$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$

Answer 2

For your problem, you should use the vector space underlying your affine space, and not think of planes, but of 2-dimensional vector spaces and (collinear) vectors which are orthogonal to them. Following John's hint, a vector orthogonal to the first (resp. second) plane is u=(2,1,1) (resp. v=(1,-1,2)). A vector orthogonal to u and v is their vector product, which is (-3,3,3); better take w=(-1,1,1). The 2-dimensional vector space orthogonal to w contains u and v by definition,so it's orthogonal to the intersection of the two given planes. Thus the plane you're looking for has an equation of the form -x+y+z=c. Since it must pass through the point (-1,0,1), the constant c must be 2 .