1The idea in one paragraph

The scalar product of two vectors is a number. The vector product, also called the cross product and written v × w, is a vector, and it is built to do one job: point at right angles to both v and w. In three dimensions two non-parallel vectors lie flat in one plane, and there is exactly one line through the origin perpendicular to that plane, so there are only two directions to choose between. The right-hand screw rule chooses one. The length is chosen to be |v||w| sin θ, which is the area of the parallelogram that v and w span, so one calculation gives you a perpendicular direction and an area at the same time. Swapping the order flips the direction. Parallel vectors span no area, so their vector product is the zero vector, which makes it a test for parallel vectors. Almost every later use of it, the normal to a plane, the direction of the line where two planes meet, is the "perpendicular to both" job.

2Calculating v × w

v × w = (v₂w₃ − v₃w₂, v₃w₁ − v₁w₃, v₁w₂ − v₂w₁). From the formula booklet.

It looks like a lot to remember, but you do not need to: it is in the booklet. What you need is a reliable way to use it. Figure 1 shows the pattern. For each component, cover that component's own row, and cross-multiply the other two rows: "down-right minus up-right", taking the rows in the cyclic order y, z for the first component, z, x for the second, and x, y for the third.

Take v = (2, −1, 3) and w = (1, 4, −2).

Always check it. The answer must be perpendicular to both vectors, so both scalar products must be zero. This takes twenty seconds and catches nearly every slip, most of them in the middle component.

Many students prefer to write it as a determinant with i, j, k across the top row, expanding along that row. That is the same calculation; the minus sign on the j term in the determinant is what produces v₃w₁ − v₁w₃. Use whichever layout you trust, and always check with the scalar product.

The base vectors. Putting i = (1, 0, 0) and j = (0, 1, 0) into the formula gives i × j = (0, 0, 1) = k. The three base vectors go round in a cycle, as Figure 4 in section 4 shows: i × j = k, j × k = i, k × i = j, and each reversed order gives the negative.

3What v × w means: direction and length

The formula is the practical definition. The geometric definition says what the answer is:

v × w = |v||w| sin θ n̂, where θ is the angle between v and w (0 ≤ θ ≤ π) and n̂ is the unit vector perpendicular to both, in the direction given by the right-hand screw rule.

Figure 2 draws it.

Direction. Imagine turning v towards w through the angle θ. A right-handed screw (an ordinary screw or bottle top) turned that way moves in the direction of v × w. Equivalently: curl the fingers of your right hand from v to w, and your thumb points along v × w. In Figure 2, v is turned anticlockwise to w as seen from above, so v × w points up.

Length. |v × w| = |v||w| sin θ. Since 0 ≤ θ ≤ π, sin θ ≥ 0, so this is never negative, as a length must be. The length is largest when v and w are perpendicular (sin θ = 1) and zero when they are parallel (sin θ = 0).

A quick numerical check of the two definitions agreeing: |(−10, 7, 9)| = √(100 + 49 + 81) = √230. Also |v| = √14, |w| = √21, and v · w = 2 − 4 − 6 = −8. The two products are linked by

That identity is not one you need to quote, but it is a good check and a neat Paper 3 "show that".

Why not use v × w to find θ? You could, from sin θ = |v × w| ÷ (|v||w|), but sin θ cannot tell an acute angle from its obtuse partner: sin 60° = sin 120°. Use the scalar product (3.13) for angles. Here cos θ = −8 ÷ √294 < 0, so θ is obtuse (about 118°), and the sine route would have wrongly suggested 62°.

4The properties

v × w = −(w × v), u × (v + w) = u × v + u × w, (kv) × w = k(v × w), v × v = 0. For non-zero vectors, v × w = 0 exactly when v and w are parallel.

Order matters: anti-commutative. Turning w towards v is the opposite twist, so the screw moves the opposite way, as Figure 3 shows. The length |w||v| sin θ is the same, so w × v is v × w reversed. Check with the numbers: w × v = (4 · 3 − (−2)(−1), (−2)(2) − (1)(3), (1)(−1) − (4)(2)) = (10, −7, −9). This is the single biggest difference from ordinary multiplication and from the scalar product, where v · w = w · v.

The base vectors show the pattern cleanly, in Figure 4.

Distributive, and scalars come out. You can expand brackets, and a number multiplying either vector can be taken outside: (3v) × w = v × (3w) = 3(v × w). But when you expand, keep the order of every product.

A vector crossed with itself is zero. The angle is 0 and sin 0 = 0; or from the formula, v₂v₃ − v₃v₂ = 0 and so on. More generally, parallel vectors give the zero vector, and for non-zero vectors that works both ways. So v × w = 0 is a test for parallel vectors: (2, −4, 6) × (−3, 6, −9) = ((−4)(−9) − (6)(6), (6)(−3) − (2)(−9), (2)(6) − (−4)(−3)) = (0, 0, 0), so the vectors are parallel. (Spotting that one is a multiple of the other, (−3, 6, −9) = −1.5(2, −4, 6), is quicker.)

Not associative. Brackets matter: (i × i) × j = 0 × j = 0, but i × (i × j) = i × k = −j. Never write u × v × w without brackets.

Using the properties in a proof. Show that (a + b) × (a − b) = −2(a × b).

Every line uses one named property, and a "show that" answer should say which. Compare numbers, where (a + b)(a − b) = a² − b²: here the "squares" vanish, and the two middle terms, which cancel for numbers, add instead, because swapping the order flips the sign.

5Areas of parallelograms and triangles

Figure 5 shows why the length is an area. Take v as the base of a parallelogram with sides v and w. The perpendicular height is |w| sin θ. Base times height is |v||w| sin θ, which is exactly |v × w|.

Area of a parallelogram with adjacent sides v and w = |v × w| (formula booklet). Area of a triangle with two sides v and w = ½|v × w|.

The triangle version is the old ½ab sin C, but with no angle to find first, which is why it is so useful in space.

Worked: a triangle in space. Find the area of the triangle with vertices A(1, 0, 2), B(3, −1, 5) and C(2, 4, 0), drawn in Figure 6.

Both vectors must start from the same vertex (AB and AC, or BA and BC). Using AB and BC also works, because the parallelogram they span has the same area, but AB and CA with an arrow each way is where mistakes creep in; start both from one corner and there is nothing to think about.

A parallelogram from three vertices. For a parallelogram ABCD (vertices in order), AB and AD are adjacent sides, and the fourth vertex satisfies AB = DC, so c = b + (d − a). Its area is |AB × AD|.

In two dimensions. Give each vector a third component of 0. For the triangle with vertices (0, 0), (5, 1) and (2, 4), the vectors are (5, 1, 0) and (2, 4, 0), and their product is (0, 0, 5 × 4 − 1 × 2) = (0, 0, 18). The area is ½ × 18 = 9.

6A vector perpendicular to two vectors

This is the job the vector product does most in the exam. To find a unit vector perpendicular to both v = (2, −1, 3) and w = (1, 4, −2):

The negative, (1/√230)(10, −7, −9), is equally correct unless a direction is specified. Any non-zero multiple of v × w is perpendicular to both, so you may simplify it before using it as a direction: (−4, 6, 2) would become (−2, 3, 1). That is what makes the vector product the standard way to find:

• a normal vector to a plane containing two given directions (3.17);
• the direction of the line where two planes meet, as the vector product of their normals (3.18);
• a direction perpendicular to two lines at once, for example to two skew lines (3.15).