Cross product
Turning a spanner makes a nut move along the axis, at right angles to the plane of your hand movement. The cross product of two vectors is a new vector at right angles to both.
The cross product a × b has magnitude |a| |b| sin θ and is perpendicular to both a and b, following the right-hand rule. It changes sign if the order is reversed, and a × a = 0. In components it is found from a 3 × 3 determinant.
Formula: a × b = determinant of [i j k ; a₁ a₂ a₃ ; b₁ b₂ b₃], a × b = −(b × a)
Rule 1: Basic products — i × j = k, j × k = i, k × i = j.
- Easy: i × j = k and j × i = −k.
- Medium: (i + j) × (j + k) = i − j + k.
- Exam-type: (2i + 3j + k) × (i − j + 2k) = 7i − 3j − 5k.
Rule 2: Properties — Use the determinant carefully.
- Easy: a × a = 0 for any vector.
- Medium: (i + 2j + 3k) × (3i + 2j + k) = −4i + 8j − 4k.
- Exam-type: If |a| = 2, |b| = 3 and θ = 30°, then |a × b| = 3.
The cross product is a vector perpendicular to both factors. Reversing the order reverses the sign.
Expanding the determinant with the wrong sign for the j term.
Cross gives a vector at a right angle.
Practice questions
- 1Find i × j.
- 2Find j × i.
- 3Find (i + j) × (j + k).
- 4Find (2i + 3j + k) × (i − j + 2k).
- 5Find a × a for a = 3i − j.
- 6Find |a × b| if |a| = 2, |b| = 3 and θ = 30°.
- 7Find (i + 2j + 3k) × (3i + 2j + k).
- 8Show that a = 2i + 4j and b = i + 2j are parallel using the cross product.
Show answers (try them yourself first)
- 1k
- 2−k
- 3i − j + k
- 47i − 3j − 5k
- 50
- 62 × 3 × ½ = 3
- 7−4i + 8j − 4k
- 8a × b has the k part 2 × 2 − 4 × 1 = 0 and no other part, so a × b = 0 and the vectors are parallel
Areas using the cross product
The area of a field shaped like a parallelogram is base times height. The magnitude of the cross product is exactly this number, so the area comes straight from the two side vectors.
The area of a parallelogram with adjacent sides a and b is |a × b|. The area of a triangle is half of it. A unit vector perpendicular to both is (a × b)/|a × b|, or its negative.
Formula: Parallelogram area = |a × b|, triangle area = ½ |a × b|
Rule 1: Area from sides — Find the cross product, then its length.
- Easy: The parallelogram with sides 3i and 4j has area 12.
- Medium: The triangle with vertices A(1, 0, 0), B(0, 1, 0), C(0, 0, 1) has AB × AC = i + j + k, so the area is √3/2.
- Exam-type: A unit vector perpendicular to i + j and j + k is ±(i − j + k)/√3.
Rule 2: Triangle from three points — Use two sides from one vertex.
- Easy: Sides i + j and j + 2k give a cross product 2i − 2j + k and a parallelogram area 3.
- Medium: For A(1, 1, 1), B(2, 3, 5), C(4, 2, 3), AB × AC = 10j − 5k and the triangle area is 5√5/2.
- Exam-type: For a = i − j and b = i + j, a × b = 2k, so |a × b| = 2.
If a × b = 0 for non-zero vectors, the two are parallel.
Forgetting the factor ½ for a triangle.
Parallelogram is the full cross product, triangle is half.
Practice questions
- 1Find the area of the parallelogram with adjacent sides 3i and 4j.
- 2Find the area of the triangle with vertices A(1, 0, 0), B(0, 1, 0), C(0, 0, 1).
- 3Find a unit vector perpendicular to both i + j and j + k.
- 4Find the area of the parallelogram with adjacent sides i + j and j + 2k.
- 5Find the area of the triangle with sides 2i and 3j.
- 6Find the area of the triangle with vertices A(1, 1, 1), B(2, 3, 5), C(4, 2, 3).
- 7What can you say about a and b if a × b = 0 and both are non-zero?
- 8Find |a × b| for a = i − j and b = i + j.
Show answers (try them yourself first)
- 112
- 2AB = −i + j, AC = −i + k, AB × AC = i + j + k, so the area is √3/2
- 3(i − j + k)/√3
- 42i − 2j + k has magnitude 3, so the area is 3
- 5½ × |2i × 3j| = ½ × 6 = 3
- 6AB × AC = 10j − 5k, magnitude 5√5, so the area is 5√5/2
- 7They are parallel (collinear)
- 8a × b = 2k, so the magnitude is 2