The scalar product gives 63.61°, which is the angle to the normal. The angle to the plane is 26.39°. Same ratio, same question, and which one you report depends on a word.
The ratio is the same in both cases. Which inverse function you apply is decided by whether the question says line-and-plane or plane-and-plane.
Write the line in parametric form and substitute straight into the plane's Cartesian equation. One unknown, one equation.
The line r = t(1, 2, 2) gives x = t, y = 2t, z = 2t. Substituting into 2x − y + 2z = 6:
2t − 2t + 4t = 6, so 4t = 6 and t = 1.5
Put that back into the line: the point is (1.5, 3, 3). Check it in the plane: 3 − 3 + 6 = 6.
Three outcomes are possible and the algebra tells you which. One solution and the line crosses the plane. No solution, when the t terms cancel and leave a false statement, and the line is parallel to the plane and outside it. Every t a solution, when the t terms cancel and leave a true statement, and the line lies in the plane.
Both start from the same scalar product. For the direction (1, 2, 2) and the normal (2, −1, 2): the product is 4, both magnitudes are 3, so the ratio is 4/9 = 0.4444.
And then they part company:
| Question | What the ratio is the cosine of | Answer |
|---|---|---|
| Line and plane | The angle to the normal, not the plane. | arcsin(0.4444) = 26.39° |
| Two planes | The angle between the normals, which IS the angle between the planes. | arccos(0.4444) = 63.61° |
For a line and a plane you need the complement. The normal sticks out at 90° to the plane, so an angle of 63.61° to the normal is 90 − 63.61 = 26.39° to the plane itself. Rather than computing the angle and subtracting, use arcsin on the ratio directly: it gives the complement in one step.
For two planes you do not. Tilting both normals by the same amount tilts both planes by the same amount, so the two angles are equal and no complement is involved. Press the button and watch the verdict change while the ratio does not.
Take the modulus of the scalar product first. A normal can be written either way round, and reversing it changes the sign and gives the obtuse partner, 116.39° instead of 63.61°. The angle wanted is always the acute one, and |d · n| forces it without any case-checking.
Two planes either are parallel, or meet in a line. They cannot meet at a single point: each plane has two dimensions of freedom and they overlap in one.
Three planes is where it gets interesting, and the guide asks for the geometrical interpretation of the solutions, which links straight back to the linear systems at AHL 1.16:
| The system has | The planes |
|---|---|
| a unique solution | meet at one point |
| infinitely many solutions | share a line, or are all the same plane |
| no solution | have no common point: parallel, or a prism |
The prism is the case nobody expects. Take:
x + y = 1, y + z = 1, x − z = 1
The normals are (1, 1, 0), (0, 1, 1) and (1, 0, −1), and no two of them are parallel: every pairwise cross product has magnitude 1.7321, nowhere near zero. So each pair of planes genuinely meets in a line, and you might expect the three lines to meet at a point.
They do not. Add the first two: x + 2y + z = 2. The third says x = 1 + z, so substituting gives 1 + z + 2y + z = 2, hence 2y + 2z = 1 and y + z = 0.5. But the second plane says y + z = 1. Contradiction, so there is no common point at all.
What the three planes form is a triangular prism: the three lines of pairwise intersection all run in direction (1, −1, 1), so they are three parallel lines, and the planes enclose an infinite triangular tube with nothing in the middle.
The quick test is the triple scalar product n₁ · (n₂ × n₃), which is 0 here. When it is zero the three normals lie in a plane and there is no unique point; when it is non-zero there is exactly one. That is one line of work and it tells you which of the three rows above you are in.
This is the one place in the vector run where the machine does the heavy lifting honestly: a three-by-three system is tedious by hand and the machine distinguishes the three outcomes clearly.
When you may use it. Analysis Paper 1 is non-calculator, so the elimination has to be secure. Paper 2 questions on three planes are exactly what the solver is for, and the interpretation is still yours.
The mark people lose. Giving the angle to the normal when the question asked for the angle to the plane. 63.61 and 26.39 are both plausible answers and the working is identical up to the last line. The habit: if the question says "plane", use arcsin; if it says "two planes", use arccos. Write which one you are using before you press the key.
Throughout: the line r = t(1, 2, 2), the plane 2x − y + 2z = 6 with normal (2, −1, 2), and a second plane with normal (1, 2, 2).
1. Find the value of t where the line meets the plane.
2. Find the angle between the line and the plane, in degrees to 2 decimal places.
3. Find the angle between the two planes, in degrees to 2 decimal places.
4. For the prism system, find the triple scalar product n₁ · (n₂ × n₃).
5. Three planes, no two parallel, and the equations have no solution. What is the arrangement?
1 markThe substitution, or the system set up.
1 markSolving it.
1 markThe point, or the statement that there is none.
1 markOn an angle question, the correct inverse function for the shapes named.
On a three-planes question the last mark is for the geometrical interpretation in words: "no solution, so the planes form a prism" earns it and "no solution" alone does not. The guide asks for the interpretation explicitly, so it will be examined.
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