Quick questions on Vector and Cartesian equations of planes in QCE Specialist Mathematics Unit 3

Writing $\mathbf{n} = (a, b, c)$ and $\mathbf{r} = (x, y, z)$, the vector equation becomes

Given three non-collinear points $A$, $B$, $C$, form two vectors in the plane, $\vec{AB}$ and $\vec{AC}$. Their cross product is normal to both, hence normal to the plane:

The perpendicular distance from a point $P_0 = (x_0, y_0, z_0)$ to the plane $ax + by + cz + d' = 0$ is

The angle between two planes equals the angle between their normals. For normals $\mathbf{n}_1$ and $\mathbf{n}_2$,

Two non-parallel planes meet in a line whose direction is $\mathbf{n}_1 \times \mathbf{n}_2$, since the line lies in both planes and so is perpendicular to both normals. Finding one common point fixes the line. To locate a point, set one variable to a convenient value (often $z = 0$) and solve the two Cartesian equations simultaneously for the other two coordinates. The point plus the direction $\mathbf{n}_1 \times \mathbf{n}_2$ then gives the vector equation of the line of intersection.

The Cartesian form $ax + by + cz = d$ packs the whole geometry into four numbers. The vector $(a, b, c)$ points perpendicular to the plane, so two planes are parallel exactly when their coefficient triples are proportional, and identical when the constants are proportional in the same ratio too. If $d = 0$ the plane passes through the origin. Dividing through by $|\mathbf{n}| = \sqrt{a^2 + b^2 + c^2}$ produces the normal form, in which the right-hand side is the signed distance from the origin to the plane, a useful check on distance problems.