A question asking whether an angle is acute needs one line, not four. The sign of the scalar product settles it, and students who know that stop computing magnitudes they do not need.
It is the same configuration as 3.12 and that is deliberate. Say so. The two vectors and the slider are identical; only the quantity being read off has changed. A class that has met the picture once can spend all its attention on the new idea.
Drag slowly through 90° and watch the bar pass through zero and change colour. The product is exactly 0 there, not a small rounding, which is worth pointing at because students expect floating-point mess and its absence is informative: the two terms cancelled algebraically.
Then ask what the sign is telling them at 60° and at 120°. Positive and negative, acute and obtuse. That is a free answer to a question they would otherwise do in four steps.
| Question | Answer |
|---|---|
| 1. p · q | 4 − 4 + 6 = 6. |
| 2. a · b | −12 + 12 = 0. |
| 3. The angle between p and q | 68.20°, or 1.1903 radians. |
| 4. p · p | 9, which is |p|². |
| 5. What a zero product means | B. Perpendicular. |
Question 4 looks trivial and is the most useful one on the page. v · v = |v|² is how nearly every vector proof begins, and a class that has met it as an exercise rather than as a formula will reach for it later.
1 markThe scalar product from the components.
1 markBoth magnitudes.
1 markcos θ written as a quotient.
1 markThe angle, in the units asked for.
The third mark is the one that protects them. Writing "cos θ = 6/16.1555" before reaching for the inverse cosine makes the division impossible to skip, and skipping it is the error that produces an error message rather than a wrong answer.
| They wrote | What happened |
|---|---|
| A vector, like (4, −4, 6) | Multiplied component by component and did not add. The word "scalar" is in the name of the thing. Worth one firm correction. |
| 14 for p · q | Lost the middle sign: (2)(−2) is −4. Pure arithmetic, but it moves the angle from 68.20° to 29.94° and nothing looks wrong. |
| 24 for a · b | Lost the first sign: (3)(−4) is −12. They then read 24/25 = 0.96 as 16.26°, so a perpendicular pair looks nearly parallel. Parallel itself would need the full 25, and the truth is 0 and 90°. |
| 16.1555 for p · q | Gave |p||q|. Those two are equal only for parallel vectors, so this is worth naming as a test they have accidentally performed. |
| An error on the calculator | Took arccos(6). The division by the magnitudes is missing, and the machine is right to refuse. |
| 21.80° on question 3 | Inverse sine instead of inverse cosine. 90 minus the right answer, which looks plausible. |
| 1.1903 on question 3 | Radians, which is correct in the wrong unit. Give it the method marks. |
| 3 for p · p | Gave |p|. The product is the square of that. |
| "There must be an error" on question 5 | Carrying the number rule that a zero product needs a zero factor. This is the misconception the page exists for and it is worth hearing out loud. |
"Why is it called scalar?" Because the answer is a scalar, a plain number, in contrast with the vector product at 3.16, whose answer is a vector. Saying that now, before 3.16, saves confusion later: the two products are named after what they produce.
"Can the angle come out over 180°?" No. The inverse cosine returns 0 to π by its range, from 3.9, and that is exactly the right range for an angle between two vectors: the angle between them is the smaller one, and it is never reflex. The restricted range of arccos is doing useful work here rather than being a nuisance.
"What if the cosine comes out bigger than 1?" Then there is an arithmetic error, and it is worth saying so as a check. The Cauchy-Schwarz inequality guarantees |v · w| ≤ |v||w|, so the quotient is always between −1 and 1. A cosine of 1.4 means a sign lost in the components.
"Does the modulus in the parallel test matter?" Yes, and it is worth one example. Opposite vectors are parallel, and then the product is negative while |v||w| is positive. Without the modulus the test would call anti-parallel vectors non-parallel, which is wrong.
| Demonstrate | The angle in one line: cos⁻¹(dotP(p,q)/(norm(p)*norm(q))). Then do it in four separate lines beside it. The one-liner is faster and the four lines are what the marks are for, and saying that explicitly stops the strongest students losing method marks. |
|---|---|
| Where they stick | Typing arccos of the product. The machine errors, and students assume a syntax problem rather than a mathematical one. Ask what the biggest a cosine can be is. |
| The check | Run dotP on the perpendicular pair and get exactly 0. Not a tiny rounding error: the two terms cancel in whole numbers, and that exactness is worth pointing out. |
The Casio has a direct Angle command that returns the angle between two vectors in one step. Show it, then tell them not to use it alone in an exam, because the marks are distributed across the working it skips.
| Step | What |
|---|---|
| 1 | The component formula, on p and q. Get 6. Insist it is one number. |
| 2 | Then a · b on the perpendicular pair. Get 0, and let the surprise land. |
| 3 | The geometric formula, |v||w| cos θ, and why the two must agree. |
| 4 | Rearrange for cos θ. The angle between p and q, in four written lines. |
| 5 | The figure. Slider through 90°, then 60° and 120°. |
| 6 | The sign table: acute, right, obtuse, from the sign alone. |
| 7 | The perpendicular and parallel tests, stated as tests they will reuse. |
| 8 | v · v = |v|², and the remark that 3.12's magnitude was a special case of this. |
Do not say "multiply the vectors". There are two products with different answers and different uses, and "multiply" belongs to neither. Say scalar product or dot product consistently, and when 3.16 arrives the distinction will already be in place.
Do not let a vector answer to a scalar product pass as careless. It is the one error here that signals a genuine misunderstanding of what the operation produces, and it will break everything from 3.14 onwards. Mark it as wrong rather than as a slip, and say why.