Vectors: magnitude, direction, the dot product and the cross product, worked on one pair, and which question wants which

In mathematics and physics a vector has a size and a direction, and in components it is a list of numbers. Two vectors can be combined in two different ways. The dot product gives a single number and answers questions about angle, projection and perpendicularity. The cross product gives a new vector perpendicular to both and answers questions about area and normals. Students lose the mark by writing the cross product as a number, or by a sign slip in its middle component. Here is one pair of vectors, every quantity computed line by line, and a table of which product each question wants.

Vectors in components, magnitude and unit vectors

A vector in three dimensions is written as an ordered triple, u = (u₁, u₂, u₃), or in the i-j-k form u₁i + u₂j + u₃k, which says the same thing. The two notations are read as the same vector by any marker. This page uses triples throughout.

The magnitude of a vector is its length, found by the Pythagorean theorem in three dimensions: |u| = √(u₁² + u₂² + u₃²). For u = (2, −1, 3):

|u| = √(2² + (−1)² + 3²) = √(4 + 1 + 9) = √14 ≈ 3.742.

A unit vector is a vector of length 1 pointing the same way. Divide by the magnitude: u/|u| = (2/√14, −1/√14, 3/√14) ≈ (0.535, −0.267, 0.802). Check it by squaring and adding the three components: 0.286 + 0.071 + 0.643 = 1.000. A unit vector carries the direction and nothing else, which is why a direction is often asked for as one.

Two things to notice before the products. A magnitude is never negative, because it is a square root of a sum of squares. And the negative sign in a component is part of the component: when you square −1 you get 1, not −1.

The dot product: a number

The dot product multiplies matching components and adds the results:

u · v = u₁v₁ + u₂v₂ + u₃v₃.

The answer is a single number, a scalar. Its geometric meaning is the second formula, which holds for the same two vectors:

u · v = |u| |v| cos θ,

where θ is the angle between them when they are drawn from the same point. Set the two formulas equal and you can find the angle from the components. Three consequences follow at once. If the dot product is zero and neither vector is zero, cos θ = 0, so the vectors are perpendicular. If the dot product is positive the angle is acute; if negative, obtuse. And the length of the projection of u onto v, meaning the shadow u casts along v's direction, is (u · v)/|v|, with a negative sign meaning the shadow points the opposite way to v.

How to find the dot product, then, is one line of arithmetic. The angle between two vectors is that line, two magnitudes and an inverse cosine.

Worked example, part 1: dot product and angle, line by line

Let u = (2, −1, 3) and v = (1, 4, −2). Find the dot product and the angle between them.

Line 1, the dot product, matching components multiplied and added: u · v = 2(1) + (−1)(4) + 3(−2) = 2 − 4 − 6 = −8.

Line 2, the magnitudes, each component squared, summed, rooted: |u| = √14 (from above); |v| = √(1² + 4² + (−2)²) = √(1 + 16 + 4) = √21 ≈ 4.583.

Line 3, the cosine, from u · v = |u||v| cos θ: cos θ = −8/(√14 · √21) = −8/√294 ≈ −8/17.146 ≈ −0.4666.

Line 4, the angle, the inverse cosine of that number: θ = cos⁻¹(−0.4666) ≈ 117.8°.

The angle is obtuse, which the negative dot product promised in line 1. That agreement is your check: a negative dot product and an acute angle cannot both be right. To one decimal place, θ ≈ 117.8°.

One more quantity from the same lines, for a question that asks how far u reaches along v: the scalar projection is (u · v)/|v| = −8/√21 ≈ −1.746. The negative sign says u leans away from v, not toward it.

The common wrong lines

Multiplying the magnitudes and forgetting the components. A student writes u · v = √14 · √21 ≈ 17.15. That number is |u||v|, which equals the dot product only when the angle is 0 and the two vectors point the same way. The dot product needs the components multiplied pair by pair, or the cosine included.

Reading a positive angle from a negative cosine. A calculator in degrees given −0.4666 returns 117.8°; a student who dropped the sign and entered 0.4666 reads 62.2°, the supplement. The sign of the dot product and the size of the angle must agree: negative means past 90°. If your angle is under 90° and your dot product is negative, one of them is wrong.

The cross product: a vector

The cross product is defined in three dimensions only. For u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃):

u × v = (u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁).

Read the pattern once and it stays. Each component uses the two subscripts that are not its own: the first component uses 2 and 3, the second uses 3 and 1, the third uses 1 and 2. In each, the first subscript goes with u, the second with v, and then the pair is swapped and subtracted. The middle component is where students go wrong, because 3 comes before 1 there, which breaks the counting order.

The result is a vector, and it is perpendicular to both u and v. Its direction follows the right-hand rule: point the fingers of your right hand along u, curl them toward v, and your thumb points along u × v. Its magnitude is

|u × v| = |u| |v| sin θ,

which is the area of the parallelogram with u and v as adjacent sides. Half of that is the area of the triangle with u and v as two sides. So the question "is the cross product a scalar" has a two-part answer: the cross product is a vector, and its magnitude is a scalar that happens to be an area. The 3×3 determinant is another way to write the same three components, and the matrices page has the determinant itself.

Worked example, part 2: the cross product, line by line

The same u = (2, −1, 3) and v = (1, 4, −2). Find u × v, check it, and find the area of the parallelogram and the triangle they span.

Line 1, the first component, subscripts 2 and 3: u₂v₃ − u₃v₂ = (−1)(−2) − (3)(4) = 2 − 12 = −10.

Line 2, the second component, subscripts 3 and 1, in that order: u₃v₁ − u₁v₃ = (3)(1) − (2)(−2) = 3 + 4 = 7.

Line 3, the third component, subscripts 1 and 2: u₁v₂ − u₂v₁ = (2)(4) − (−1)(1) = 8 + 1 = 9.

Line 4, the vector: u × v = (−10, 7, 9).

Line 5, the perpendicularity check, a dot product with each original vector: u · (−10, 7, 9) = 2(−10) + (−1)(7) + 3(9) = −20 − 7 + 27 = 0. v · (−10, 7, 9) = 1(−10) + 4(7) + (−2)(9) = −10 + 28 − 18 = 0. Both zero, so the product is perpendicular to both, as it must be.

Line 6, the area, the magnitude of the cross product: |u × v| = √((−10)² + 7² + 9²) = √(100 + 49 + 81) = √230 ≈ 15.17. The parallelogram has area about 15.17 square units; the triangle has half that, about 7.58.

A check on line 6 from part 1: |u||v| sin θ = √14 · √21 · sin(117.8°) ≈ 17.146 × 0.8845 ≈ 15.17. The two routes agree.

The common wrong lines

Writing a single number as the cross product. A student computes 15.17 and writes "u × v = 15.17". The cross product is the vector (−10, 7, 9); 15.17 is its magnitude. Questions that ask for a normal, or for a vector perpendicular to both, want the triple; questions that ask for an area want the number. Read which one the question wants before you write the answer.

The sign in the middle. A student computes the second component in counting order, u₁v₃ − u₃v₁ = 2(−2) − 3(1) = −4 − 3 = −7, and writes (−10, −7, 9). It looks right. The check catches it: u · (−10, −7, 9) = −20 + 7 + 27 = 14, not 0. One wrong sign and the "perpendicular" vector is not perpendicular. That is why line 5 is taught as part of the method and not as an optional extra. Studying a worked example beside the problem it solves is what Sweller and Cooper (1985) found teaches a procedure like this faster than unguided attempts, and the check is the part of the example worth copying.

Siblings to try cold

Try these with the page closed, then compare under the check heading. Attempt first: it is chapter 1 of the free playbook, and testing yourself before you reread is why the attempt is the learning.

Let a = (1, 2, 2) and b = (3, 0, −4). Find |a|, |b|, a · b, the angle between a and b, a × b, and the area of the parallelogram they span.

Check

|a| = 3. |b| = 5. a · b = −5. cos θ = −5/15 = −1/3, so θ ≈ 109.5°, obtuse, as the negative dot product says. a × b = (−8, 10, −6), which dots to zero with both a and b. |a × b| = √200 = 10√2 ≈ 14.14. The numbers were chosen so that the magnitudes and the cosine need no calculator; if you got −1/3 and an angle under 90°, reread the second wrong line above.

Which question wants which product: cross product vs dot product

The question's words tell you which product it wants. Match the words to the row.

The question asks for	The product
The angle between two vectors	Dot (then the inverse cosine)
Whether two vectors are perpendicular	Dot (zero means yes)
The projection or component of one vector along another	Dot (divided by a magnitude)
Work done by a force along a displacement	Dot
A vector perpendicular to two others, or a normal to a plane	Cross
The area of a parallelogram or a triangle	Cross (its magnitude)
Torque	Cross

One sentence covers the rest: the dot product exists in any number of dimensions, because it only multiplies and adds; the cross product, as defined here, works only in three, because only there do two vectors leave exactly one perpendicular direction.

Where this sits in the course, and where it goes next

In the MCV4U course these are three ideas in a row: idea 19, Cartesian vectors in two and three dimensions with magnitude, direction and unit vectors; idea 20, the dot product, the angle between vectors and projections; idea 21, the cross product with area and applications. Before them is idea 18, vectors as geometric and algebraic objects with addition and scalar multiplication. After them comes idea 22, the vector, parametric and scalar equations of lines, and the strand ends with planes and intersections. The MCV4U course map places the vector strand against the calculus half of the course.

The vector strand has no counterpart in first-year calculus; at university it belongs to linear algebra, where the dot product is idea 5, length, dot product and angle, of 26.

What the two products are used for next: the normal vector of a plane (MCV4U idea 23) is a cross product of two vectors in the plane; the distance from a point to a line (idea 24) uses both products; and in physics the dot product is work and the cross product is torque. When you practise, work one pair of vectors through both products as this page did, then do a mixed set where the question does not say which product it wants: worked examples, then practice problems has the order.

The two products on one screen

• Magnitude: |u| = √(u₁² + u₂² + u₃²); a unit vector is u/|u|.
• Dot product: u · v = u₁v₁ + u₂v₂ + u₃v₃ = |u||v| cos θ. A number. Zero means perpendicular; negative means obtuse.
• Cross product: u × v = (u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁). A vector, perpendicular to both; |u × v| = |u||v| sin θ is the parallelogram's area.
• The check: dot the cross product with each original vector; both must be zero. A nonzero result is almost always the middle component's sign.
• Which product: angle, perpendicularity, projection, work → dot; normal, area, torque → cross.
• Where: the dot product in any dimension; the cross product in three only.