☰ خطة العام · الرياضيات البحتة (بالإنجليزية)

The vector product, areas and the scalar triple product

الأسبوع 11 · الضرب الاتجاهي: التعريف والإحداثيات والتوازي والمساحات والضرب الثلاثي القياسي والحجوم · 3 دروس
2

The vector product in coordinates (a 3 by 3 determinant), its properties and the test for parallel vectors

Pure Mathematics Sec 3 Term 1, Unit 3, Lesson 3
GoalCalculate A × B from components with a 3 by 3 determinant, use its properties, and test whether two vectors are parallel.
New words
determinant · a number made from a square table of numbers; for 2 by 2: |a b; c d| = ad − bcA × B in coordinates · (AᵧB_z − AzBᵧ, AzBₓ − AₓB_z, AₓBᵧ − AᵧBₓ)test for parallel vectors · A × B is the zero vector, or one vector is a multiple of the other
Explanation

The book writes A × B as a 3 by 3 determinant, so we learn that first (in Arabic المحدد). A 2 by 2 determinant is |a b; c d| = ad − bc: multiply along the main diagonal, subtract the product along the other diagonal. To expand a 3 by 3 determinant along its first row, take each number of the first row, multiply it by the small 2 by 2 determinant that is left when you cross out its row and its column, and give the three terms the signs + − +. The middle sign is minus: that is the usual place for mistakes. Now write A = (Aₓ, Aᵧ, Az) and B = (Bₓ, Bᵧ, Bz). Then A × B is the determinant whose first row is i, j, k, whose second row is A and whose third row is B. Expanding gives A × B = (AᵧB_z − AzBᵧ) i − (AₓB_z − AzBₓ) j + (AₓBᵧ − AᵧBₓ) k. The table shows where each part comes from. For vectors of the plane, Az = Bz = 0 and only the k part is left: A × B = (AₓBᵧ − AᵧBₓ) k. Properties: A × B is perpendicular to both A and B (always check this with two dot products, as it catches slips); A × B = −B × A; A × (B + C) = A × B + A × C; (kA) × B = k(A × B). Parallel vectors: A × B is the zero vector exactly when A and B are parallel (or one of them is zero), that is, exactly when the components are proportional, A = kB.

cross out the row of i, j, k and the column of that letter; the middle sign is minus
part of A × Bsmall 2 × 2 determinantsign
i partAᵧB_z − AzBᵧ+
j partAₓB_z − AzBₓ−
k partAₓBᵧ − AᵧBₓ+
Worked examples
A = (2, −1, 3), B = (1, 4, −2). The i part is (−1)(−2) − 3 × 4 = −10. The j part is −[2 × (−2) − 3 × 1] = 7. The k part is 2 × 4 − (−1) × 1 = 9. So A × B = (−10, 7, 9). Check: A · (A × B) = −20 − 7 + 27 = 0 and B · (A × B) = −10 + 28 − 18 = 0. Reversing the order gives B × A = (10, −7, −9).
Find a unit vector perpendicular to the plane of A = (1, 2, 2) and B = (2, 1, −2). A × B = (2 × (−2) − 2 × 1, −[1 × (−2) − 2 × 2], 1 × 1 − 2 × 2) = (−6, 6, −3). Its size is √(36 + 36 + 9) = 9, so the unit vector is (−6, 6, −3)/9 = (−2/3, 2/3, −1/3). The opposite vector (2/3, −2/3, 1/3) is also a correct answer.
Plane vectors: (3, 2) × (−1, 4) = (3 × 4 − 2 × (−1)) k = 14k. The answer points along the z-axis, out of the plane of the two vectors.
Parallel vectors. (2, −6, 4) and (−3, 9, −6) are parallel: the second is −3/2 times the first, and the cross product is the zero vector. For which k and m is (k, −4, 6) parallel to (3, 2, m)? Write (k, −4, 6) = λ(3, 2, m). From −4 = 2λ we get λ = −2, so k = −6 and 6 = −2m gives m = −3.
Class activity

Make a cross-product card. Write the three parts of the formula on a card as in the table, and practise on the vectors (1, 0, 2) and (0, 1, −3): the answer is (−2, 3, 1). Then test it with two dot products, which must both be 0.

Practice
1
Find (1, 3, 0) × (2, −1, 4).
2
Find (4, 0, −2) × (1, 1, 3) and check that it is perpendicular to (4, 0, −2).
3
Show that (−2, 6, −4) is parallel to (1, −3, 2), by the cross product.
4
Find m if (−4, 10, 6) is parallel to (2, m, −3).
5
Find the vector product of the plane vectors (5, −2) and (3, 1).