A plane equation built from the wrong three numbers is a perfectly well-formed equation for the wrong plane. One scalar product catches it, and it is worth making that check a reflex before 3.18 arrives.
Press to the third candidate, the one labelled "the coefficients", and ask whether it lies in the plane. A good fraction of any class will say yes, because the numbers came off the equation. The readout says 9, and the arrow is drawn standing straight up out of the plane.
The figure draws each candidate at its true angle to the normal, computed from the scalar product, so the first two lie flat, the third is vertical, and the fourth sits at 48 degrees. It is not a schematic; the picture is the arithmetic.
Then ask what the first two candidates have in common. Both give zero, and both lie in the plane, which is the statement worth collecting: v · n = 0 means v is in the plane.
| Question | Answer |
|---|---|
| 1. The constant d | a · n = 6. |
| 2. (1, 2, 0) · n | 0, so it is in the plane. |
| 3. n · n | 9, so the normal is not. |
| 4. The expression at (1, 2, 2) | 4, not 6, so the point is off the plane. |
| 5. What the coefficients are | B. The normal. |
Questions 2 and 3 are the whole lesson as a pair: one vector gives zero and lies in the plane, the other gives 9 and does not. Both are three numbers taken from the same equation, which is exactly why the confusion exists.
1 markThe normal, found or quoted.
1 markThe constant, from a · n.
1 markThe equation in the required form.
When the question supplies two in-plane directions, the first mark is the cross product and carries most of the work. Insist the normal is written on its own line: it banks a mark even if the constant then goes wrong, and it makes the checking step possible.
| They wrote | What happened |
|---|---|
| A plane using an in-plane direction as the normal | The error this page exists for. Both are three numbers. The one-line check catches it; nothing else does. |
| 0 for n · n | Assumed the normal is perpendicular to itself, usually by over-applying "a vector in the plane gives zero". No non-zero vector is perpendicular to itself. |
| 10 for d | Lost the middle sign: (2)(−1) is −2. Arithmetic, but it produces a parallel plane in the wrong place. |
| 9 for d | Dotted the normal with itself instead of with the point. Worth asking which vector they used, because the two questions look alike on the page. |
| 3 for a scalar product | Added the components instead of multiplying matching pairs. Back to 3.13. |
| A parametric form with one parameter | They have written a line, not a plane. A plane is two-dimensional and needs two parameters; worth saying as a dimension count. |
| Two parallel vectors in the parametric form | Then the two directions span a line rather than a plane, and the cross product comes out as the zero vector. That vanishing cross product is the diagnosis. |
| Forgetting d entirely | Writing 2x − y + 2z = 0 describes a parallel plane through the origin. Worth noting that d = 0 has that specific meaning rather than being a missing number. |
"Why does one normal fix a whole plane?" Count dimensions. A plane in three dimensions has two directions of freedom, and one perpendicular direction uses up the third. So specifying the perpendicular is just as informative as specifying the two directions, and it is one vector instead of two. That argument also explains why a line in three dimensions needs a direction rather than a normal: there is no single perpendicular to a line in space.
"Can I scale the normal?" Yes, and the equation scales with it. 4x − 2y + 4z = 12 is the same plane as 2x − y + 2z = 6, with the normal doubled and d doubled too. Students who multiply through and then compare with a book's answer need to know that, and it is the same non-uniqueness as 3.14, so it is worth linking.
"How do I find two vectors in the plane from the Cartesian form?" Find any two points on it and subtract, or find any two vectors perpendicular to the normal by inspection. For (2, −1, 2), (1, 2, 0) works because 2 − 2 + 0 = 0. Tell them it does not have to be elegant and there is no unique answer; finding one by trial is a legitimate method.
"Is the parametric form examined?" It is in the content list, so yes, but questions overwhelmingly use the Cartesian and scalar product forms, because they are shorter and because 3.18 needs them. Teach all three and spend the time on the conversions.
| Demonstrate | The whole plane in two lines: crossP(b,c) for the normal, then dotP(a,crossP(b,c)) for the constant. Two keystrokes from a point and two directions to 2x − y + 2z = 6, which makes the structure of the method obvious. |
|---|---|
| Where they stick | Expecting the machine to produce an equation. It produces the two ingredients and the assembling is theirs. Say so, because the third mark is for exactly that step. |
| The check | dotP(b,crossP(b,c)) returning 0. Make it routine here as well as at 3.16, because a wrong normal is invisible in the final equation. |
Testing a point needs no vector work at all: evaluate the expression in Run-Matrix and compare with d. Worth showing, because students go looking for the matrix and vector editor when plain arithmetic on the same screen would do it.
| Step | What |
|---|---|
| 1 | A plane needs two directions, or one normal. Count the dimensions out loud. |
| 2 | r · n = a · n from a point and a normal. Multiply out to the Cartesian form. |
| 3 | Notice the coefficients ARE the normal. Then ask whether the normal is in the plane. |
| 4 | The figure, straight to the third candidate. |
| 5 | The two statements: v · n = 0 means in the plane; v parallel to n means perpendicular to it. |
| 6 | Parametric to Cartesian via the cross product, with the perpendicularity check. |
| 7 | Testing points by substitution, and the d = 0 remark about the origin. |
| 8 | The one-line check to apply to every plane equation they ever write. |
Do not say "the coefficients give the direction of the plane". A plane has no single direction, and the sentence points students at exactly the wrong interpretation. Say the coefficients are the normal, and add that the normal is perpendicular to the plane, in the same breath, every time.
Do not let a plane equation be written without the check. It costs one line and it catches the one error on this sub-topic that is otherwise undetectable. By 3.18 a wrong normal will produce wrong angles and wrong intersections with no symptom at all, so the habit has to be in place before then.