The cross product of two vectors is a third vector perpendicular to both; it exists only in three dimensions. Its magnitude is the area of the parallelogram the two span. A plane is fixed by a point and a normal vector ⟨a, b, c⟩: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0. The coefficients are the normal.
Step 1: Picture It
Start with something you already know.
Pushing a door open
- A door turns on its hinge, a straight line up its edge.
- Push level with the floor and square to the door, and it swings easily; push straight at the hinge, and nothing turns.
- Push at a slant, and only part of your push does any turning.
- The arm is the line across the door from the hinge to your hand.
- When your push is level with the floor, the hinge stands at right angles to both the arm and your push.
The cross product of the door’s arm and your push points along the hinge: at right angles to both of them.
Its size is the turning effect: largest when the push is square to the arm, and zero when the push runs along it.
Where the picture falls short: the hinge is only a line, but a cross product picks one direction along it, and swapping the order flips that direction.
The same idea with numbers. An arm of length 2 and a push of size 3, at different angles θ. The size of the cross product is 2 × 3 × sin θ.
| Angle between arm and push | Size of the cross product |
|---|---|
| 0° | 0 |
| 30° | 3 |
| 90° | 6 |
| 150° | 3 |
| 180° | 0 |
The size is |u||v| sin θ: zero when the two run along one line, and the full |u||v| at a right angle. It is also the area of the parallelogram the two vectors span.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
u = ⟨1, 2, 0⟩ and v = ⟨0, 1, 1⟩. Using only dot products, which of these vectors is at right angles to both u and v?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Computing a cross product
u × v = ⟨u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁⟩. For ⟨1, 0, 0⟩ × ⟨0, 1, 0⟩: ⟨0 − 0, 0 − 0, 1 − 0⟩ = ⟨0, 0, 1⟩. i × j = k, as the right hand says.
A worked one
u = ⟨1, 2, 3⟩, v = ⟨4, 5, 6⟩: ⟨2·6 − 3·5, 3·4 − 1·6, 1·5 − 2·4⟩ = ⟨−3, 6, −3⟩. Check perpendicularity: ⟨−3, 6, −3⟩ · ⟨1, 2, 3⟩ = −3 + 12 − 9 = 0. And · v: −12 + 30 − 18 = 0.
Order matters
v × u = ⟨3, −6, 3⟩, the exact opposite of u × v. Swapping the order reverses the result, which never happens with a dot product. Parallel vectors give the zero vector.
Its length is an area
|⟨−3, 6, −3⟩| = √(9 + 36 + 9) = √54 ≈ 7.35: the area of the parallelogram spanned by u and v. Half of that, 3.67, is the area of the triangle with those two sides.
A plane through three points
P(1, 0, 0), Q(0, 2, 0), R(0, 0, 3): PQ = ⟨−1, 2, 0⟩, PR = ⟨−1, 0, 3⟩. PQ × PR = ⟨6, 3, 2⟩, the normal. Plane: 6(x − 1) + 3y + 2z = 0, so 6x + 3y + 2z = 6. Check: Q gives 6, R gives 6.
Reading a plane equation
3x − 2y + z = 7 has normal ⟨3, −2, 1⟩, already written down. Two planes are parallel when their normals are parallel; the angle between planes is the angle between their normals.
Step 3: Try It Yourself
Tap and try it out.
Find the plane through P(1, 0, 0), Q(0, 2, 0) and R(0, 0, 3), and the area of triangle PQR
- 1PQ = ⟨−1, 2, 0⟩, PR = ⟨−1, 0, 3⟩end minus start, twice
Step 0 of 4
- i-hat lands on(3, 1)
- j-hat lands on(1, 2)
- Determinant5
The shaded parallelogram is the image of the unit square, and its area is 5. That is exactly what the determinant measures.
Step 4: In Real Life
A wrench and a roof
Torque is the cross product of the wrench’s arm and the force: a vector along the bolt, whose size is the turning effect. Three points on a roof define its plane through a cross product.
Step 5: Watch an Example
One step at a time.
Watch Ana Cross Two Vectors and Measure the Area
Ana wants u × v for u = ⟨1, 1, 0⟩ and v = ⟨0, 1, 2⟩, then the area the two vectors span.
- Step 1
Ana lines u = ⟨1, 1, 0⟩ up above v = ⟨0, 1, 2⟩. Each component of u × v is built from the other two columns, never its own: the first component uses the second and third entries, the second uses the third and first, and the third uses the first and second.
Watch Leo Build a Plane, Then Finish It Yourself
Leo wants the plane through P(1, 2, 0), Q(3, 3, 0) and R(2, 3, 4), and the area of triangle PQR. He makes the edges; the normal, the equation and the area are yours.
- Step 1
Leo makes two edges from the same corner, end minus start: PQ = Q − P = ⟨2, 1, 0⟩ and PR = R − P = ⟨1, 1, 4⟩. Both lie in the plane, so anything perpendicular to both of them is a normal.
Step 6: Your Turn
Practice makes it stick.
The Shade Sail
Problem 1 of 14
A triangular shade sail is tied to three posts at A(0, 0, 2), B(4, 0, 2) and C(0, 3, 6), measured in meters. The fabric costs $12 per square meter. What does the fabric for the sail cost, in dollars?
dollars
The Sloping Roof
Problem 2 of 14
A flat roof passes through three corners P(0, 0, 3), Q(6, 0, 3) and R(0, 4, 5), in meters, where z is the height above the ground. How high is the roof directly above the point x = 3, y = 2, in meters?
meters
Cross and Plane
Work each one a line at a time. A dot check catches most slips in a cross product.
Problem 3 of 14
u = ⟨3, 0, 1⟩ and v = ⟨1, 2, 0⟩. What is the first component of u × v?
Problem 4 of 14
What is the area of the parallelogram spanned by u = ⟨3, 0, −1⟩ and v = ⟨0, 2, −1⟩, in square units?
square units
Problem 5 of 14
A plane passes through (2, −1, 3) with normal ⟨1, 4, −2⟩. Written as x + 4y − 2z = d, what is d?
Problem 6 of 14
u × v = ⟨2, −5, 1⟩. What is the second component of v × u?
Problem 7 of 14
Two sides of a triangle are 8 cm and 5 cm long, with an angle of 30° between them. What is the triangle’s area, in square centimeters?
square centimeters
Whose Step?
One line on each page is a slip. Find it, then put it right.
Problem 8 of 14
Dev worked out ⟨1, 3, 2⟩ × ⟨2, 1, 0⟩ one component at a time. One line is a slip. Which?
Dev’s working for ⟨1, 3, 2⟩ × ⟨2, 1, 0⟩
- 1first: 3·0 − 2·1 = −2
- 2second: 1·0 − 2·2 = −4
- 3third: 1·1 − 3·2 = −5
- Answeru × v = ⟨−2, −4, −5⟩
Problem 9 of 14
Worked from the formula, what is the second component of ⟨1, 3, 2⟩ × ⟨2, 1, 0⟩?
Problem 10 of 14
Maya wrote the plane through (3, −2, 1) with normal ⟨2, 1, −1⟩. One line is a slip. Which?
Maya’s plane through (3, −2, 1) with normal ⟨2, 1, −1⟩
- 12(x − 3) + 1(y − 2) − 1(z − 1) = 0
- 22x − 6 + y − 2 − z + 1 = 0
- Answer2x + y − z = 7
Problem 11 of 14
With the sign put right, the plane through (3, −2, 1) with normal ⟨2, 1, −1⟩ reads 2x + y − z = d. What is d?
Pick the Move
Choose the move that does the job. Neither needs working all the way through.
Problem 12 of 14
A support rod must point at right angles to both u = ⟨2, 0, 1⟩ and v = ⟨1, 3, 0⟩. Which move gives such a direction in one go?
Problem 13 of 14
Three corners of a flat solar panel are P, Q and R. Which move comes first on the way to the equation of the panel’s plane?
Same Plane?
Judge each one against the normal and a point. Select every entry that fits.
Problem 14 of 14
Select every equation that describes the same plane as 3x + y − 2z = 6.
Step 7: Quick Check
Show what you know.
Question 1 of 2
Two vectors of length 4 and 6 meet at 30 degrees. What is the cross product magnitude?
Question 2 of 2
What is a normal vector to a plane?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
For which number k do the four points A(0, 0, 0), B(1, 1, 0), C(0, 1, 1) and D(2, k, 1) all lie in one plane?
Stretch 2 of 2
A triangle has corners O(0, 0, 0), A(2, 0, 0) and B(0, b, 3), with b positive. Its area is 5 square units. What is b?
What You Learned
- The cross product gives a vector perpendicular to both inputs, reverses when the order swaps, and exists only in three dimensions.
- Its magnitude is the area of the parallelogram the two vectors span; half of it is a triangle.
- A plane is fixed by a point and a normal, and the normal’s components become the coefficients.
- Three points give a plane: two edge vectors, their cross product, then the point-normal form.
For the grown-up
The misconception this lesson targets is that the cross product is the dot product with a different symbol: multiply matching components, and the order does not matter. A student holding it multiplies u₁v₁ for the first component and gets a vector that is not perpendicular to anything in particular, or writes v × u and u × v as the same thing, so a normal comes out pointing the wrong way without anyone noticing. The slips that travel with it are a component subtracted in the reverse order and a sign dropped in y − y₀ when the point has a negative coordinate. The cheap habit that catches all of them is the dot check: a true cross product dots to zero with both of its inputs, and a true plane equation is satisfied by every point it was built from.