For 2x − y + 2z = 6 the vector (2, −1, 2) is the normal. Dot it with itself and you get 9, not 0, so it is nowhere near lying in the plane. The numbers in front of x, y and z point straight out.
The plane is drawn edge on, so it looks like a line. A vector in the plane lies flat along it; the normal stands straight up out of it. Those are the only two cases that matter.
| Form | Written | Use it when you have |
|---|---|---|
| Parametric | r = a + sb + tc | A point and two non-parallel vectors in the plane. |
| Scalar product | r · n = a · n | A point and the normal. |
| Cartesian | ax + by + cz = d | Anything. It is the scalar product form multiplied out. |
A plane needs two directions, because it is two-dimensional, which is why the parametric form has two parameters. Or it needs one normal, because a single perpendicular direction pins a plane down completely. Those are two ways of saying the same thing, and the normal version is almost always shorter.
For the plane through A(1, 2, 3) with normal n = (2, −1, 2):
r · n = a · n = (1)(2) + (2)(−1) + (3)(2) = 6
and writing r = (x, y, z) and multiplying out gives 2x − y + 2z = 6. The coefficients are the normal's components and the right-hand side is a · n.
So the coefficients are not a direction in the plane. Press through the candidates and watch the readout. (1, 2, 0) gives 0 and lies flat in the plane. (0, 2, 1) gives 0 as well. But (2, −1, 2), which is exactly the list of coefficients, gives 9, and a vector in the plane must give zero.
That is a useful thing to be able to say the right way round:
Reading the coefficients as a direction is the error that makes a line come out perpendicular to a plane when the question wanted it lying in one, and nothing about the answer looks wrong.
Parametric to Cartesian: cross the two directions. Given the plane r = (1, 2, 3) + s(1, 2, 0) + t(0, 2, 1), the two directions are in the plane, so the normal is perpendicular to both, which is exactly what a vector product gives:
(1, 2, 0) × (0, 2, 1) = (2, −1, 2)
and then r · n = 6 as before. The cross product from 3.16 is the bridge between the two forms, which is the main reason that sub-topic comes first. Check it the usual way: (2, −1, 2) dotted with each of the two directions gives 0.
Cartesian to parametric: read the normal off, then find two vectors perpendicular to it. That is easier than it sounds, because you only need any two that work, and they do not have to be tidy.
Testing whether a point is on a plane is one substitution. For 2x − y + 2z = 6:
| Point | 2x − y + 2z | On the plane? |
|---|---|---|
| (1, 2, 3) | 6 | yes |
| (2, 4, 3) | 6 | yes |
| (3, 4, 2) | 6 | yes |
| (1, 2, 2) | 4 | no |
| (0, 0, 0) | 0 | no |
The last row is worth noticing: the origin is on the plane exactly when d = 0. So a Cartesian equation with no constant term describes a plane through the origin, which is a free piece of information about any such equation.
The one real computation here is the cross product that turns two in-plane directions into a normal, and that is worth checking because the component pattern is fiddly. Everything else is one substitution.
When you may use it. Analysis Paper 1 is non-calculator, and plane work is nearly all Paper 1: the normals and the constants are whole numbers by design. The machine is for confirming a cross product.
The mark people lose. Using a direction in the plane where the normal was needed. Both are three numbers, both look like vectors, and the resulting equation is a perfectly well-formed plane equation for the wrong plane. The habit: after writing any plane equation, dot its coefficients with a direction you know is in the plane and check you get zero. One line, and it catches the whole class of error.
Throughout: the plane through A(1, 2, 3) with normal n = (2, −1, 2).
1. Find the constant d in 2x − y + 2z = d.
2. Find (1, 2, 0) · n.
3. Find n · n.
4. Evaluate 2x − y + 2z at the point (1, 2, 2).
5. What do the coefficients 2, −1 and 2 in 2x − y + 2z = 6 represent?
1 markThe normal, found or quoted correctly.
1 markThe constant, from a · n.
1 markThe equation in the form the question asked for.
If the question gives two directions in the plane, the first mark is the cross product and it is worth two of the three marks' worth of effort. Write the normal down on its own line before assembling the equation, so it earns its mark even if the constant goes wrong.
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