☰ Year plan · Pure Mathematics

The vector product, areas and the scalar triple product

Week 11 Β· The vector product: definition, coordinates, parallel vectors, areas, the scalar triple product and volumes Β· 3 lessons
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 Γ— B from components with a 3 by 3 determinant, use its properties, and test whether two vectors are parallel.
New words
determinant Β· a number made from a square table of numbers; for 2 by 2: |a b; c d| = ad βˆ’ bcA Γ— B in coordinates Β· (AᡧB_z βˆ’ AzBᡧ, AzBβ‚“ βˆ’ Aβ‚“B_z, Aβ‚“Bᡧ βˆ’ AᡧBβ‚“)test for parallel vectors Β· A Γ— B is the zero vector, or one vector is a multiple of the other
Explanation

The book writes A Γ— B as a 3 by 3 determinant, so we learn that first (in Arabic Ψ§Ω„Ω…Ψ­Ψ―Ψ―). A 2 by 2 determinant is |a b; c d| = ad βˆ’ 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 + βˆ’ +. The middle sign is minus: that is the usual place for mistakes. Now write A = (Aβ‚“, Aᡧ, Az) and B = (Bβ‚“, Bᡧ, Bz). Then A Γ— 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 Γ— B = (AᡧB_z βˆ’ AzBᡧ) i βˆ’ (Aβ‚“B_z βˆ’ AzBβ‚“) j + (Aβ‚“Bᡧ βˆ’ AᡧBβ‚“) 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 Γ— B = (Aβ‚“Bᡧ βˆ’ AᡧBβ‚“) k. Properties: A Γ— B is perpendicular to both A and B (always check this with two dot products, as it catches slips); A Γ— B = βˆ’B Γ— A; A Γ— (B + C) = A Γ— B + A Γ— C; (kA) Γ— B = k(A Γ— B). Parallel vectors: A Γ— 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 Γ— Bsmall 2 Γ— 2 determinantsign
i partAᡧB_z βˆ’ AzBᡧ+
j partAβ‚“B_z βˆ’ AzBβ‚“βˆ’
k partAβ‚“Bᡧ βˆ’ AᡧBβ‚“+
Worked examples
A = (2, βˆ’1, 3), B = (1, 4, βˆ’2). The i part is (βˆ’1)(βˆ’2) βˆ’ 3 Γ— 4 = βˆ’10. The j part is βˆ’[2 Γ— (βˆ’2) βˆ’ 3 Γ— 1] = 7. The k part is 2 Γ— 4 βˆ’ (βˆ’1) Γ— 1 = 9. So A Γ— B = (βˆ’10, 7, 9). Check: A Β· (A Γ— B) = βˆ’20 βˆ’ 7 + 27 = 0 and B Β· (A Γ— B) = βˆ’10 + 28 βˆ’ 18 = 0. Reversing the order gives B Γ— A = (10, βˆ’7, βˆ’9).
Find a unit vector perpendicular to the plane of A = (1, 2, 2) and B = (2, 1, βˆ’2). A Γ— B = (2 Γ— (βˆ’2) βˆ’ 2 Γ— 1, βˆ’[1 Γ— (βˆ’2) βˆ’ 2 Γ— 2], 1 Γ— 1 βˆ’ 2 Γ— 2) = (βˆ’6, 6, βˆ’3). Its size is √(36 + 36 + 9) = 9, so the unit vector is (βˆ’6, 6, βˆ’3)/9 = (βˆ’2/3, 2/3, βˆ’1/3). The opposite vector (2/3, βˆ’2/3, 1/3) is also a correct answer.
Plane vectors: (3, 2) Γ— (βˆ’1, 4) = (3 Γ— 4 βˆ’ 2 Γ— (βˆ’1)) k = 14k. The answer points along the z-axis, out of the plane of the two vectors.
Parallel vectors. (2, βˆ’6, 4) and (βˆ’3, 9, βˆ’6) are parallel: the second is βˆ’3/2 times the first, and the cross product is the zero vector. For which k and m is (k, βˆ’4, 6) parallel to (3, 2, m)? Write (k, βˆ’4, 6) = Ξ»(3, 2, m). From βˆ’4 = 2Ξ» we get Ξ» = βˆ’2, so k = βˆ’6 and 6 = βˆ’2m gives m = βˆ’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, βˆ’3): the answer is (βˆ’2, 3, 1). Then test it with two dot products, which must both be 0.

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