One scalar product ratio answers two different questions, and the answers are 63.61° and 26.39°. Both are plausible, the working is identical until the last line, and only the word "plane" in the question tells you which.
Press between the two views and watch the ratio readout not change. 0.4444 in both. Then watch the verdict change from arcsin to arccos. That pairing is the lesson: the arithmetic is shared and the interpretation is not.
On the first view, point at the two arcs and at the little right-angle square on the normal. The normal is at 90° to the plane, so the two marked angles must add to 90. That is the whole reason a complement appears, and it is a picture rather than a rule.
On the second view there is no right angle to borrow: both normals tilt with their planes, so the two marked angles are equal. Ask the class why the complement has disappeared, and the figure answers it.
| Question | Answer |
|---|---|
| 1. t at the intersection | 4t = 6, so t = 1.5, point (1.5, 3, 3). |
| 2. Line and plane | arcsin(0.4444) = 26.39°. |
| 3. Two planes | arccos(0.4444) = 63.61°. |
| 4. The triple scalar product | 0. |
| 5. The arrangement | B. A triangular prism. |
Questions 2 and 3 are the same ratio and the same calculator keystrokes up to the final function. Marking them together makes the diagnosis immediate: a student who swaps the two answers has understood everything except which shape was named.
1 markThe substitution, or the system set up.
1 markSolving it.
1 markThe point, or a statement that there is none.
1 markOn an angle question, the right inverse function for the shapes named.
On three planes the final mark is for words. The guide asks for the geometrical interpretation of the solutions explicitly, so "no solution, so the three planes form a prism" earns it and "no solution" does not. Tell them the guide says so; it is more persuasive than being told it is good practice.
| They wrote | What happened |
|---|---|
| 63.61° for a line and a plane | The error this page exists for. Gave the angle to the normal. Press between the two figure views rather than explaining. |
| 26.39° for two planes | The same error inverted: took a complement where none was needed. Both halves of the confusion appear in any class. |
| 116.39° | Normal written the other way round, giving the obtuse partner. The modulus on the scalar product prevents it entirely. |
| 3 for t | Gave a coordinate of the point rather than the parameter. Same error as 3.15 question 3, and worth linking. |
| −1 for the triple product | Stopped after the first of the three terms. The three are −1, +1 and 0. |
| 1.7321 for the triple product | Gave a pairwise cross product's magnitude. That answers "are any two parallel", which is a different and also necessary question. |
| "Impossible" on question 5 | Carrying the two-dimensional intuition that non-parallel things must meet. The explicit system settles it, and it is worth doing the three lines of algebra on the board. |
| "No solution" with no interpretation | Three of four marks. The sentence is the fourth. |
"Why does the line case need a complement and the plane case not?" Because of what the normal is attached to. A line's direction lies along the line, so the angle to the normal is 90 minus the angle to the plane. A plane's normal is perpendicular to the plane, and both planes' normals are, so tilting the planes tilts the normals equally and the two angles coincide. Count the right angles: one in the first case, two in the second, and two cancel.
"How do I know whether a line lies in a plane?" Substitute and watch what happens to the parameter. If the t terms cancel and leave something false, the line is parallel and outside. If they cancel and leave something true, every t works and the line lies in the plane. That is three outcomes from one substitution, and students who expect only "find t" are thrown by the other two.
"Is the triple scalar product on the course?" Not as named content, which is why the page offers it as a quick test rather than a method. The examinable route is solving the system and interpreting the solutions, which links to AHL 1.16. Offer the triple product as a check and make clear the marks are for the system.
"What are all the three-plane arrangements?" Worth listing once: one point; a common line; all three coincident; two parallel with a third cutting both; all three parallel; and the prism. The last is the only one students do not predict, which is why it is the one the page dwells on.
| Demonstrate | Put the prism system in and get no solution. Then change the third plane from x-z=1 to x-z=0: the three planes now share the whole line r = (0, 1, 0) + t(1, −1, 1). One digit apart, and two completely different geometries, which makes the interpretation feel consequential rather than decorative. Note it does NOT become a unique point, and it cannot: the triple scalar product is 0 either way, so the coefficient matrix is singular and changing a right-hand side can only move you between "none" and "infinitely many". |
|---|---|
| Where they stick | Expecting the machine to classify the three cases for them. It cannot: a singular coefficient matrix stops the SIMUL solver, and "no solution" and "infinitely many" both have a zero determinant, so the solver cannot tell them apart. The geometry is theirs to read off the algebra, which is exactly where the last mark is. |
| The check | The triple scalar product, DotP(VctA,CrossP(VctB,VctC)), returning 0. It confirms the solver's verdict by a completely different route, which is worth more than repeating the same calculation. |
For the angles, put the modulus into the expression once and for all: abs(dotP(d,n)). It removes the obtuse answer as a possibility rather than requiring a check afterwards.
| Step | What |
|---|---|
| 1 | A line meeting a plane, by substitution. One unknown, one equation, t = 1.5. |
| 2 | The other two outcomes: parallel outside, and lying in the plane. All three from the same substitution. |
| 3 | The angle to the normal, from 3.13. Then ask what the question actually wanted. |
| 4 | The figure, first view. The right-angle square, the two arcs, the sum of 90. |
| 5 | Second view. Two normals, two right angles, no complement. |
| 6 | Two planes meet in a line, never a point. Count the dimensions. |
| 7 | Three planes and the three outcomes, linked to AHL 1.16. |
| 8 | The prism, worked explicitly: y + z = 0.5 against y + z = 1. |
Do not say "use the scalar product to find the angle". It is the instruction that produces 63.61° on a line-and-plane question. Say "the scalar product gives the angle to the normal", every time, and then ask what the question wanted. The longer sentence is the whole content of the sub-topic.
Do not let "no solution" stand as a complete answer. It is algebra, and the guide asks for geometry. A class that writes the interpretation every time will not lose that mark, and a class that treats it as optional will lose it in the paper, where it is often the last and most valuable mark on a long question.