What is the difference between dot product and cross product?

The dot product a · b = |a||b| cos θ is a number and tells you about angles and perpendicularity. The cross product a × b is a vector perpendicular to both a and b whose length |a||b| sin θ equals the area of the parallelogram they span.

How do I know my cross product is correct?

Dot your answer with each of the original vectors. Both results must be 0. If either is not, there is an arithmetic error, most often the sign of the j component.

Which way does a × b point?

It follows the right-hand rule: curl the fingers of your right hand from a towards b and your thumb points along a × b. Seeing it in a rotatable 3D graph makes the direction much easier to judge than a flat diagram.

Dot Product and Cross Product (DSE M2)

The scalar product a · b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃ is a number. The vector product a × b is a vector perpendicular to both a and b, with magnitude |a||b| sin θ, found by expanding a 3 × 3 determinant.

Solve Scalar Product and Vector Product questions with AI

a × b = i − 7j − 5k from Example 1, standing perpendicular to the plane of a and b. Try it yourself →

Dot Product and Cross Product Formulas

Scalar producta · b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃

Perpendicular testa · b = 0 ⟺ a ⊥ b (non-zero vectors)

Self producta · a = |a|²

Vector product (determinant form)a × b = (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k

Magnitude of a × b|a × b| = |a||b| sin θ

Order mattersb × a = −(a × b), a × a = 0

Link between the two|a × b|² = |a|²|b|² − (a · b)²

How to Calculate a Cross Product with a Determinant

Write i, j, k in the first row, the components of a in the second row and of b in the third.
Expand along the first row. Cover the column of i and take the 2 × 2 determinant of what is left; repeat for j and k.
Use the signs + − +: the j term is subtracted.
Check: take the dot product of your answer with a and with b. Both must be 0, because a × b is perpendicular to each of them. This takes ten seconds and catches almost every arithmetic slip.

Worked Examples — Scalar Product and Vector Product

Example 1: Finding a × b with a determinant

Let a = 2i + j − k and b = i − 2j + 3k. Find a × b and verify that it is perpendicular to both a and b.

i component: (1)(3) − (−1)(−2) = 1
j component: −[(2)(3) − (−1)(1)] = −7
k component: (2)(−2) − (1)(1) = −5
a × b = i − 7j − 5k
Check: (1)(2) + (−7)(1) + (−5)(−1) = 0 and (1)(1) + (−7)(−2) + (−5)(3) = 0 ✓

Answer: a × b = i − 7j − 5k

Example 2: Angle between two vectors with the dot product

Find the angle between a = 3i + 4k and b = i + 2j + 2k, correct to 3 significant figures.

a · b = (3)(1) + (0)(2) + (4)(2) = 11
|a| = √(9 + 16) = 5, |b| = √(1 + 4 + 4) = 3
cos θ = 11 / (5 × 3) = 11/15
θ ≈ 42.8°

Answer: θ ≈ 42.8°

💡 Exam Traps & Secrets

⚠️ The j term has a minus signThe most common cross-product slip is forgetting the minus in front of the j determinant. Always run the two dot-product checks before moving on.

⚠️ a × b and b × a are oppositeThe cross product is not commutative: b × a = −(a × b). Keep the order the question gives. The dot product, by contrast, gives a number, so never write it in i, j, k form.

⚠️ Scalar triple product is no longer requiredThe current curriculum notes say the scalar triple product and the term "parallelepiped" are not required. Older past papers still use them, so do not be alarmed if you meet one, but they are not something to drill.

✏️ Self-Test: Quick Interactive Practice

How to study: Click any question below to instantly load it on the coordinate grid and check your steps.

(Cross product)Find a × b for a = i + 2j + 3k and b = 2i − j + k, and check it is perpendicular to both.

(Dot product = 0)Find the value of k if 2i + kj − k and i − j + 4k are perpendicular.

(Angle between vectors)Find the angle between 2i − j + 2k and i + j.