C. Geometry and Algebra of Vectors · Dot and Cross ProductsMagnitude of the Cross Product: Areas of Parallelograms and Triangles
C2.6 C2.8About 70 minutes
Learning goals- I can explain why |u × v| = |u||v| sin θ.
- I can find the area of a parallelogram or triangle in two-space or three-space using the cross product.
- I can use the cross product to test whether two vectors are parallel.
- I can choose between the dot product and the cross product for a given task.
Words to know
magnitude of the cross product
|u × v| = |u||v| sin θ, the area of the parallelogram formed by u and v.
Lagrange's identity
|u × v|² = |u|²|v|² − (u · v)².
collinear vectors
Vectors that are parallel (they lie along the same line when placed tail to tail).
parallelogram determined by two vectors
The parallelogram with the two vectors as adjacent sides starting from one corner.
How long is the cross product?
In lesson 7 you learned the direction of u × v: perpendicular to both vectors, by the right-hand rule. Its length has a beautiful meaning too:
|u × v| = |u||v| sin θ
where θ is the angle between u and v (tail to tail, 0° ≤ θ ≤ 180°). Compare with the dot product, u · v = |u||v| cos θ: one uses cosine, the other sine.
Why it is true. By expanding both sides in components (try it in the Challenge), you can show Lagrange's identity:
|u × v|² = |u|²|v|² − (u · v)²
Now replace u · v with |u||v| cos θ: |u × v|² = |u|²|v|² − |u|²|v|² cos²θ = |u|²|v|²(1 − cos²θ) = |u|²|v|² sin²θ. Since sin θ ≥ 0 for 0° ≤ θ ≤ 180°, taking square roots gives |u × v| = |u||v| sin θ.
Area of a parallelogram and a triangle
Place u and v tail to tail and complete the parallelogram. Use |u| as the base. The height is the perpendicular distance from the tip of v to the base, which is |v| sin θ. So
area of parallelogram = |u||v| sin θ = |u × v|
and the triangle with sides u and v is half of the parallelogram:
area of triangle = ½|u × v|
Written with trigonometry instead of vectors, this is the formula area = ½ab sin C for a triangle with sides a and b and contained angle C: the height on side a is b sin C, so ½ × base × height = ½ab sin C. (You may not have met this formula in MCR3U; it follows straight from ½ × base × height.) The vector form has a big advantage: it works for triangles anywhere in three-space, straight from the coordinates of the vertices.
Worked exampleExample 1: From lengths and an angle
|u| = 6, |v| = 4 and the angle between them is 30°. Find |u × v| and the area of the triangle they form.
- |u × v| = (6)(4) sin 30° = 24 × 1/2 = 12.
- Triangle: ½ × 12 = 6.
|u × v| = 12 (the parallelogram's area) and the triangle's area is 6 square units.
Worked exampleExample 2: A parallelogram from components
Find the area of the parallelogram determined by u = [2, −1, 3] and v = [1, 4, −2]. Check with Lagrange's identity.
- From lesson 7, u × v = [−10, 7, 9].
- |u × v| = √(100 + 49 + 81) = √230 ≈ 15.17.
- Check: |u|² = 4 + 1 + 9 = 14, |v|² = 1 + 16 + 4 = 21 and u · v = 2 − 4 − 6 = −8. Then |u|²|v|² − (u · v)² = 294 − 64 = 230 ✓.
Area = √230 ≈ 15.17 square units.
Worked exampleExample 3: A triangle in three-space
Find the area of the triangle with vertices A(1, 0, 2), B(3, 4, 1) and C(0, 2, 5).
- Use two sides from the same vertex: AB = [2, 4, −1] and AC = [−1, 2, 3].
- AB × AC = [(4)(3) − (−1)(2), (−1)(−1) − (2)(3), (2)(2) − (4)(−1)] = [14, −5, 8].
- |AB × AC| = √(196 + 25 + 64) = √285.
- Area = ½√285 ≈ 8.44.
About 8.44 square units. (In lesson 4 you found ∠A ≈ 79.9° for this triangle; ½|AB||AC| sin A = ½√21 √14 sin 79.9° ≈ 8.44 as well ✓.)
Worked exampleExample 4: A triangle in two-space
Find the area of the triangle with vertices P(1, 1), Q(6, 2) and R(3, 5).
- PQ = [5, 1] and PR = [2, 4]. Treat them as [5, 1, 0] and [2, 4, 0].
- From lesson 7, [a, b, 0] × [c, d, 0] = [0, 0, ad − bc]. Here ad − bc = (5)(4) − (1)(2) = 18.
- Area = ½|18| = 9.
9 square units. In two-space, the area of the parallelogram on [a, b] and [c, d] is simply |ad − bc|.
Parallel vectors
If u and v are parallel (θ = 0° or 180°), then sin θ = 0, so u × v = 0. The parallelogram has collapsed into a line segment with no area. Conversely, if u × v = 0 for non-zero vectors, they must be parallel. This is a quick collinearity test.
Worked exampleExample 5: A parallel test
Are [2, −4, 6] and [−1, 2, −3] parallel?
- [2, −4, 6] × [−1, 2, −3] = [(−4)(−3) − (6)(2), (6)(−1) − (2)(−3), (2)(2) − (−4)(−1)] = [12 − 12, −6 + 6, 4 − 4] = [0, 0, 0].
Yes. (Indeed [2, −4, 6] = −2[−1, 2, −3]: they point in opposite directions.)
On 0° to 180°, sine is positive and takes most values twice, so |u × v| = |u||v| sin θ cannot tell an acute angle from its obtuse partner. The dot product (cosine) can.Which product for which job?
Use the dot product to find angles. If you use |u × v| = |u||v| sin θ, then sin⁻¹ only gives angles up to 90°, and you cannot tell θ from 180° − θ. Use the cross product for areas, perpendicular directions and parallel tests.
Just for funA parallelogram walks into a geometry class and says, "Don't worry about my area, my cross product has it covered." The triangle next to it says, "I only need half of that."
Hands-on activityA leaning deck of cards
You need: a deck of playing cards (or a pad of sticky notes, or a stack of about 50 sheets of paper), a ruler and a protractor.
- Stand the deck on its long edge on the table and look at it from the side. Its side face is a rectangle: base b (the card length) and height h (the deck's thickness, now vertical).
- Push the top of the deck sideways so the cards slide over each other. The side face becomes a parallelogram. Measure the slanted side length |v| and the angle θ between the slanted side and the base.
- Compute b × |v| sin θ, which is |u × v| with |u| = b. Compare with b × h for the upright deck.
- Lean the deck at two other angles and repeat.
What to look for: the slanted side gets longer as you lean the deck, but |v| sin θ stays equal to the height h, so the area stays the same each time (within measurement error). For an 8.9 cm card and a 1.6 cm thick deck, the area stays close to 14 cm².
Your turn
1|u| = 5, |v| = 8 and θ = 150°. Find |u × v|.
40 sin 150° = 40 × 1/2 = 20.
2Find the area of the parallelogram determined by [1, 2, 3] and [4, 5, 6].
Cross product [−3, 6, −3]; area √(9 + 36 + 9) = √54 = 3√6 ≈ 7.35.
3Find the area of the triangle determined by [3, 0, −1] and [2, 1, 4].
Cross product [1, −14, 3]; area ½√(1 + 196 + 9) = ½√206 ≈ 7.18.
4Find the area of the parallelogram determined by [4, 1] and [2, 3] in two-space.
|ad − bc| = |12 − 2| = 10.
5Find the area of the triangle with vertices O(0, 0, 0), A(2, 0, 0) and B(0, 3, 0) two ways.
Cross product: [2, 0, 0] × [0, 3, 0] = [0, 0, 6], area ½(6) = 3. Right-triangle check: ½(2)(3) = 3 ✓.
6Use the cross product to decide whether [3, −6, 9] and [−1, 2, −3] are parallel.
Cross: [(−6)(−3) − (9)(2), (9)(−1) − (3)(−3), (3)(2) − (−6)(−1)] = [0, 0, 0]: parallel (the first is −3 times the second).
More practice: warm-up
1|u| = 3, |v| = 7 and u ⊥ v. Find |u × v|.
2u and v point in the same direction. Find |u × v|.
3What is the area of the square determined by i and j?
4Write the formula for the area of a triangle with sides u and v.
No. It is a length, so it is never negative.
More practice: on your own
1Find the area of the triangle with vertices K(1, 1, 0), L(2, 3, 1) and M(0, 2, 3).
KL = [1, 2, 1], KM = [−1, 1, 3]. Cross: [6 − 1, −1 − 3, 1 + 2] = [5, −4, 3], length √50 = 5√2. Area = 5√2/2 ≈ 3.54.
2ABCD is a parallelogram with A(2, 1, −1), B(5, 2, 1) and D(3, −1, 2). Find C and the area.
AB = [3, 1, 2], AD = [1, −2, 3]. C = B + AD = (6, 0, 4). AB × AD = [3 + 4, 2 − 9, −6 − 1] = [7, −7, −7]. Area = 7√3 ≈ 12.12.
3For the parallelogram in the previous question, find the height measured from base AB.
Height = area ÷ base = 7√3 ÷ √14 ≈ 3.24.
4|u| = 4, |v| = 6 and |u × v| = 12. What can you say about θ? What extra information would settle it?
sin θ = 12/24 = 1/2, so θ = 30° or θ = 150°. The sign of u · v would settle it: positive means 30°, negative means 150°.
5Find the area of the triangle with vertices (−2, 1), (3, 4) and (1, −3).
Vectors from (−2, 1): [5, 3] and [3, −4]. ad − bc = −20 − 9 = −29. Area = ½|−29| = 14.5.
Challenge
1Prove Lagrange's identity |u × v|² = |u|²|v|² − (u · v)².
Left side: (u₂v₃ − u₃v₂)² + (u₃v₁ − u₁v₃)² + (u₁v₂ − u₂v₁)². Right side: (u₁² + u₂² + u₃²)(v₁² + v₂² + v₃²) − (u₁v₁ + u₂v₂ + u₃v₃)². On the right, the terms u₁²v₁², u₂²v₂² and u₃²v₃² cancel, leaving u₁²v₂² + u₂²v₁² − 2u₁v₁u₂v₂ = (u₁v₂ − u₂v₁)², plus the two similar groups for the pairs (2, 3) and (3, 1). These are exactly the three squares on the left. ∎
2Find the area of the triangle with vertices (1, 0, 0), (0, 2, 0) and (0, 0, 3).
Vectors from (1, 0, 0): [−1, 2, 0] and [−1, 0, 3]. Cross product [6, 3, 2], length √(36 + 9 + 4) = 7. Area = 3.5.
3Find all k for which the triangle with vertices O(0, 0, 0), A(1, 2, 0) and B(0, k, 1) has area 1.5.
[1, 2, 0] × [0, k, 1] = [2, −1, k]. Area ½√(5 + k²) = 1.5, so 5 + k² = 9 and k = 2 or k = −2.
Quiz
1. |u| = 4, |v| = 5 and the angle between them is 30°. What is |u × v|?
2. What is the area of the parallelogram determined by [2, 0, 0] and [0, 3, 0]?
3. What is the area of the triangle with vertices (0, 0, 0), (4, 0, 0) and (0, 3, 0)?
4. u and v are non-zero and u × v = 0. What must be true?
5. Why is the dot product, not the cross product, the right tool for finding the angle between two vectors?
6. What is the area of the parallelogram determined by [1, 2, 3] and [4, 5, 6]?
Big idea (C2.6, C2.8): the length of the cross product is |u × v| = |u||v| sin θ, which equals the area of the parallelogram formed by u and v. Half of it is the area of the triangle. If the cross product is the zero vector, the vectors are parallel.
How to help: The deck-of-cards activity is quick and shows why area depends on height, not on the slanted side. Ask your teen to explain why ½|u × v| gives the same result as ½ × base × height, which in trigonometric form is ½ab sin C.
Common difficulties:
- Forgetting the ½ for triangles.
- Worrying about which sides to use. Any two different sides of the triangle, written as vectors, give the same area (AB × AC and AB × BC have the same length); the usual habit is two sides leaving one vertex.
- Trying to find an obtuse angle with sin⁻¹. Use the dot product for angles.
- Arithmetic slips inside the square root; check each component of the cross product with dot products first.