Angle between Two Planes

Consider two planes $\vec r\cdot\vec n_1=d_1$ and $\vec r\cdot\vec n_2=d_2$, with normals $\vec n_1$ and $\vec n_2$. The inclination of the two planes to each other is completely decided by the inclination of $\vec n_1$ and $\vec n_2$ — if the normals are perpendicular, the planes are perpendicular too. So the planes are perpendicular to each other if and only if $\vec n_1\cdot\vec n_2=0$. In Cartesian form, for planes $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$, the normals are $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$, so this perpendicularity condition becomes $a_1a_2+b_1b_2+c_1c_2=0$.

When the two planes are not perpendicular, "the angle between the planes" is defined, by convention, as the acute angle between their normals — two intersecting planes actually form a pair of supplementary angles at their line of intersection, and by convention the acute one is always reported. Using the dot-product formula for the angle between two vectors, and taking the modulus of the cosine to force an acute result:

Worked example. Find the angle between the planes $\vec r\cdot(\hat i+\hat j-2\hat k)=8$ and $\vec r\cdot(-2\hat i+\hat j+\hat k)=3$.

Here $\vec n_1=\hat i+\hat j-2\hat k$ and $\vec n_2=-2\hat i+\hat j+\hat k$.

$\vec n_1\cdot\vec n_2=(1)(-2)+(1)(1)+(-2)(1)=-2+1-2=-3$.

$|\vec n_1|=\sqrt{1+1+4}=\sqrt6$ and $|\vec n_2|=\sqrt{4+1+1}=\sqrt6$. …