The vector, scalar product and Cartesian equations of a plane, the normal vector, and the intersection of a line with a plane and of two planes.

What this dot point is asking

OCR wants you to write the equation of a plane in vector form, in scalar product form $\mathbf{r}\cdot\mathbf{n} = d$, and in Cartesian form $ax + by + cz = d$, to find the normal vector (directly or from the cross product of two vectors in the plane), and to find the intersection of a line with a plane (a point) and of two planes (a line).

The normal vector and the scalar product form

The key object for a plane is its normal, a vector perpendicular to every line in the plane. The scalar product form says that the displacement from a fixed point in the plane to any other point is perpendicular to the normal.

The coefficients in the Cartesian equation are the components of the normal, so you can read the normal straight off any Cartesian plane equation, which is used in every angle and distance calculation.

Finding the normal from points in the plane

If you are given three points, or a point and two directions lying in the plane, the normal is the cross product of two such direction vectors, because the cross product is perpendicular to both.

Line meets plane

A line generally pierces a plane at a single point. Parametrise the line, substitute its coordinates into the plane equation, and solve the resulting equation for the single unknown $\lambda$.

Plane meets plane

Two non-parallel planes intersect in a line. Solving the two Cartesian equations simultaneously leaves one free parameter, which traces the line of intersection: set one coordinate equal to a parameter $\lambda$, solve the two equations for the other two coordinates in terms of $\lambda$, and read off the result as a vector equation $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$. The direction $\mathbf{d}$ of that line is perpendicular to both normals, so an alternative route is to compute $\mathbf{n}_1\times\mathbf{n}_2$ for the direction and then find one point on both planes. If instead the normals are parallel, the planes are parallel: they either coincide (the same equation up to a scalar) or never meet, which is exactly the singular matrix picture of two of the three planes in a linear system.

Planes pull together the cross product (for the normal), the scalar product (for $d$ and for angles), and the matrix view of intersecting planes.

Try this

Q1. State the normal vector of the plane $4x - y + 2z = 7$. [1 mark]

• Cue. The coefficients give $\mathbf{n} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k}$.

Q2. Does the point $(1, 1, 1)$ lie in the plane $x + 2y - z = 2$? [1 mark]

• Cue. $1 + 2 - 1 = 2$, which equals the right-hand side, so yes.