Source: Homework Set 1, §12.4
Tags: cross product, determinant, normal vector, area of triangle, area of parallelogram, right-hand rule, vector product
Difficulty: Intermediate | Prerequisites: §12.1-12.3 (vectors, dot product, projection). Familiarity with 2x2 and 3x3 determinants is helpful but not required.
Big Picture
The cross product is the second major vector operation in Calculus III (the dot product being the first). Unlike the dot product, which returns a scalar, the cross product returns a new vector that is perpendicular to both inputs. This makes it the natural tool for finding normal vectors to planes, computing areas of triangles and parallelograms, and later for calculating torque and flux. §12.4 introduces the cross product and its geometric meaning; the homework tests both computational skill and conceptual understanding (such as when a cross product equation can or cannot be solved).
The cross product a x b gives a vector perpendicular to both a and b, with magnitude equal to the area of the parallelogram they span. You compute it using a 3x3 determinant. A cross product equation v x b = c has a solution only when c is perpendicular to b, and when solutions exist, they are not unique.
Cross product (a x b)
A binary operation on two vectors in R³ that produces a third vector perpendicular to both. Defined via the determinant of a 3x3 matrix with i, j, k in the first row.
Think of it as: the operation that answers "which direction is perpendicular to both of these?"
Right-hand rule
The convention for the direction of a x b: curl the fingers of your right hand from a toward b, and your thumb points in the direction of a x b.
In simple terms, it tells you which of the two possible perpendicular directions the cross product picks.
Normal vector
A vector perpendicular to a surface or plane. The cross product of two vectors lying in a plane gives a normal to that plane.
Parallelogram area
The magnitude |a x b| equals the area of the parallelogram formed by a and b. Half of this gives the area of the triangle they span.
In simple terms, the cross product's length measures how much area the two vectors enclose.
Computing the cross product
Set up a 3x3 determinant with i, j, k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row.
Expand along the first row using cofactors. The result is a new vector.
Example: ⟨a₁, a₂, a₃⟩ x ⟨b₁, b₂, b₃⟩ = ⟨a₂b₃ - a₃b₂, a₃b₁ - a₁b₃, a₁b₂ - a₂b₁⟩.
Key properties
Anti-commutativity: a x b = -(b x a). Swapping the order flips the sign.
Orthogonality: (a x b) · a = 0 and (a x b) · b = 0. The result is always perpendicular to both inputs.
Scalar multiplication: (ca) x b = c(a x b).
Distributivity: a x (b + c) = a x b + a x c.
a x a = 0 for any vector a.
When does v x b = c have a solution?
A necessary condition is that c must be perpendicular to b, that is, c · b = 0. This follows from the orthogonality property: any cross product with b is perpendicular to b.
To show no solution exists (as in Q2b), check whether c · b = 0. If it is not zero, no such v exists.
When a solution does exist, it is not unique, because adding any scalar multiple of b to v does not change v x b.
Worked example: finding v such that v x ⟨5, -2, 6⟩ = ⟨2, 2, -1⟩ (Q2a)
First check: ⟨2, 2, -1⟩ · ⟨5, -2, 6⟩ = 10 - 4 - 6 = 0. The condition is satisfied, so solutions exist.
Set v = ⟨x, y, z⟩ and expand v x ⟨5, -2, 6⟩. Equate components to ⟨2, 2, -1⟩ and solve the resulting system. One solution is sufficient; the answer is not unique.
Worked example: showing no v exists for v x ⟨5, -2, 6⟩ = ⟨1, -1, -1⟩ (Q2b)
Check: ⟨1, -1, -1⟩ · ⟨5, -2, 6⟩ = 5 + 2 - 6 = 1 ≠ 0.
Since the right-hand side is not perpendicular to ⟨5, -2, 6⟩, no vector v can produce it as a cross product with ⟨5, -2, 6⟩.
Area of triangle T from a cube (Q2c)
Place the cube with vertex a at the origin and side length 6. Identify the coordinates of p and q as midpoints of their respective edges.
Form vectors from a to p and from a to q.
The area of triangle T = (1/2)|(a to p) x (a to q)|.
Compute the cross product, take its magnitude, and halve it.
\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}Expanded:
\mathbf{a} \times \mathbf{b} = \langle a_2 b_3 - a_3 b_2,\; a_3 b_1 - a_1 b_3,\; a_1 b_2 - a_2 b_1 \rangleArea of the parallelogram spanned by a and b:
\text{Area}_{\text{parallelogram}} = |\mathbf{a} \times \mathbf{b}|Area of the triangle spanned by a and b:
\text{Area}_{\text{triangle}} = \tfrac{1}{2}\,|\mathbf{a} \times \mathbf{b}|Torque is defined as the cross product of the position vector and the force vector: τ = r x F. The direction of the torque tells you the axis of rotation, and the magnitude tells you how much rotational force is applied.
In physics and engineering, the cross product also appears in the magnetic force on a charged particle: F = q(v x B), where v is velocity and B is the magnetic field.
Students often treat the cross product as commutative. It is not: a x b = -(b x a). Swapping the order reverses the direction.
Confusing the cross product (a vector) with the dot product (a scalar). If your answer to a cross product is a single number, something has gone wrong.
Forgetting that the cross product is only defined in R³. There is no cross product of two vectors in R².
When solving v x b = c, students sometimes assume the solution is unique. It is not: you can always add any multiple of b to a valid solution without changing the cross product.
⚠️ Computing cross products by determinant expansion is a bread-and-butter exam skill. Practise until it is automatic.
⚠️ The condition for a cross product equation v x b = c to have a solution (c · b = 0) is a common "show that" problem. Know why it must hold.
⚠️ Area-of-triangle problems using the cross product appear frequently. Remember the factor of 1/2.
⚠️ Anti-commutativity (a x b ≠ b x a) is a favourite exam trap, especially in multi-step problems.
True or false: a x b = b x a.
False. a x b = -(b x a).
Fill in the blank: |a x b| gives the area of the ___ spanned by a and b.
Parallelogram.
True or false: if a x b = 0, then a and b are parallel.
True (assuming both are nonzero).
Fill in the blank: a cross product equation v x b = c can only be solved if c · b = ___.
0
True or false: the cross product is defined for vectors in R².
False. It is only defined in R³.
Q: Compute ⟨1, 2, 3⟩ x ⟨4, 5, 6⟩.
A: ⟨(2)(6) - (3)(5), (3)(4) - (1)(6), (1)(5) - (2)(4)⟩ = ⟨12 - 15, 12 - 6, 5 - 8⟩ = ⟨-3, 6, -3⟩.
Q: What is the area of the parallelogram spanned by ⟨1, 0, 2⟩ and ⟨0, 3, 1⟩?
A: Cross product = ⟨(0)(1) - (2)(3), (2)(0) - (1)(1), (1)(3) - (0)(0)⟩ = ⟨-6, -1, 3⟩. Magnitude = sqrt(36 + 1 + 9) = sqrt(46). The area is sqrt(46).
Q: Why is there no vector v with v x ⟨1, 0, 0⟩ = ⟨1, 1, 0⟩?
A: Check ⟨1, 1, 0⟩ · ⟨1, 0, 0⟩ = 1 ≠ 0. Since the right-hand side is not perpendicular to ⟨1, 0, 0⟩, no solution exists.
Q: If a x b = ⟨2, -1, 3⟩, what is b x a?
A: b x a = -(a x b) = ⟨-2, 1, -3⟩.
Q: Two sides of a triangle are given by vectors ⟨3, 0, 0⟩ and ⟨0, 4, 0⟩. What is the area of the triangle?
A: Cross product = ⟨0, 0, 12⟩. Magnitude = 12. Triangle area = 12/2 = 6.
The cross product is the tool you use to find normal vectors to planes (§12.5). When you have two direction vectors lying in a plane, their cross product gives the normal, which you then use to write the plane equation.
It connects to orthogonal projections (§12.2-12.3) through the rotation algorithm in Question 5, where the normal to the rotation plane is defined as v₁ x v₂.
In physics courses, the cross product reappears in angular momentum, torque, and electromagnetism.
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