a = (3, 4) and b = (−4, 3) are both 5 long, and a · b = −12 + 12 = 0. Nothing here is zero except the answer, and that is the most useful fact in the whole vector run.
Both vectors keep their length of 5 throughout. Only the angle changes, and the product follows it from +25 to −25.
The scalar product has a component formula and a geometric one, and the whole usefulness of it is that both are true at once:
v · w = v1w1 + v2w2 + v3w3
v · w = |v| |w| cos θ
The first is arithmetic you can do without thinking. The second contains the angle. Set them equal and you have the angle between any two vectors, which is the one technique this sub-topic exists to give you:
cos θ = (v · w) / (|v| |w|)
For p = (1, 2, 2) and q = (4, −2, 3):
The answer is a scalar, which is what the name says. If your answer to a scalar product has brackets round it, you have multiplied component by component and kept them separate, which gives (4, −4, 6). Add them up.
Because |v| and |w| are never negative, the sign of v · w is the sign of cos θ. So:
| v · w | cos θ | The angle |
|---|---|---|
| positive | positive | acute, under 90° |
| zero | zero | exactly 90° |
| negative | negative | obtuse, over 90° |
So you can answer "is this angle acute or obtuse" without finding it at all: compute the product and look at the sign. That is a one-line answer to a question that otherwise needs three steps.
The perpendicular test, and why it is surprising. With ordinary numbers, a product of zero means one of them was zero. With vectors it does not:
a · b = (3)(−4) + (4)(3) = −12 + 12 = 0
and both vectors have length 5. The two terms cancelled. So:
for non-zero vectors, v · w = 0 means exactly that they are perpendicular
This is the test you will use most often in the rest of the topic: to check a normal to a plane, to find a perpendicular direction, to confirm a right angle in a solid. Move the slider to 90° and the readout reaches zero exactly, not nearly.
And the parallel test. At the other extreme, cos θ is ±1, so:
for parallel vectors, |v · w| = |v| |w|
Check it: p · 2p = 18, and |p| |2p| = 3 × 6 = 18. Equal, so parallel. For p and q, p · q = 6 against |p||q| = 16.1555, nowhere near, so not parallel.
The modulus on the left matters: two vectors pointing in opposite directions are still parallel, and then the product is negative while the magnitudes' product is positive. Without the modulus the test would reject them.
| Property | Checked on p and q |
|---|---|
| v · w = w · v | Both 6. Order does not matter. |
| u · (v + w) = u · v + u · w | p · (q + p) = 15, and 6 + 9 = 15. |
| (kv) · w = k(v · w) | (3p) · q = 18, and 3 × 6 = 18. |
| v · v = |v|² | p · p = 9, and |p|² = 3² = 9. |
The last one is worth more than it looks. It says the magnitude is not a separate idea: |v| = √(v · v), and the whole of 3.12 is a special case of this sub-topic. It is also how most vector proofs get started, because squaring a magnitude turns a length into something you can expand.
Both machines have a dot product command, which turns the angle calculation into three lines. The value is in checking a component sum you have already done, because a single sign error in the components changes the angle completely.
When you may use it. Analysis Paper 1 is non-calculator, and these questions are usually built so the numbers are kind: 6 over 16.1555 is a Paper 2 answer, and Paper 1 will give you something whose cosine is a neat fraction.
The mark people lose. Taking the inverse cosine of the scalar product itself. arccos(6) does not exist, so the machine errors and students assume they have mistyped. The product must be divided by the two magnitudes first, and the result has to land between −1 and 1. The habit: write cos θ = ... before you write θ = ..., so the division cannot be skipped.
Throughout: p = (1, 2, 2), q = (4, −2, 3), a = (3, 4, 0) and b = (−4, 3, 0).
1. Find p · q.
2. Find a · b.
3. Find the angle between p and q, in degrees to 2 decimal places.
4. Find p · p.
5. a · b = 0 and neither vector is zero. What does that tell you?
1 markThe scalar product from the components.
1 markBoth magnitudes.
1 markcos θ as a quotient, written down.
1 markThe angle, in the units asked for.
Four marks and only the last one needs the calculator. A question asking merely whether an angle is acute needs only the first mark's work, so read the question before computing magnitudes you may not need.
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