Multiply matching pairs and add them up: Picture two arrows. The scalar (dot) product v·w squeezes them into a single number (a scalar, not a vector).
There are two equal recipes
- Components: multiply the x's, multiply the y's (and z's), then add.
- Magnitudes & angle: |v| × |w| × cos θ, where θ is the angle between them.
Pick the right one: The IB gives you both formulas — picking the right one is the whole skill.
IB-style question — dot product from components
Vectors are given by p = (3, −2, 4) and q = (1, 5, −2).
Find p·q.
Step by step
- Multiply each matching pair of components.
- Work out each product.
- Add them to get a single number.
Final answer
p·q = −15 (a scalar, not a vector).
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Set the two formulas equal, then solve for cos θ: Both recipes give the same number, so you can set them equal and solve for the angle θ.
Finding the angle
Set them equal
v·w (from components) = |v||w| cos θ.
Make cos θ the subject
Rearrange so cos θ is on its own.
Take cos⁻¹
Inverse cosine gives θ.
Magnitude: The magnitude of a vector is |v| = √(v₁² + v₂² + v₃²) — the length of the arrow (from Pythagoras).
IB-style question — angle between two vectors
Two vectors are u = (1, 2, 2) and v = (2, 0, −1).
Find the angle θ between u and v, giving your answer in degrees.
Step by step
- Top of the fraction — the dot product.
- The magnitudes (lengths) of each vector.
- Put them into cos θ.
- Take cos⁻¹ of 0.
Final answer
θ = 90° (the vectors are perpendicular, since the dot product is 0).
IB-style question — angle that isn't a right angle
Find the angle between a = (3, 4) and b = (5, 0), to the nearest degree.
Step by step
- Dot product (2-D, so just two terms).
- Magnitudes.
- cos θ.
- Inverse cosine.
Final answer
θ ≈ 53° (1 d.p.: 53.1°).