# Cross product and determinants (Sect. 12.4) Two main ways to introduce the cross product

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1 Cross product and determinants (Sect. 12.4) Two main ways to introduce the cross product Geometrical definition Properties Expression in components. Definition in components Properties Geometrical expression. We choose the first way, like the textbook.

2 Cross product and determinants (Sect. 12.4) Geometric definition of cross product Definition The cross product of vectors v and w in R 3 having magnitudes v, w and angle in between θ, where 0 θ π, is denoted by v w and is the vector perpendicular to both v and w, pointing in the direction given by the right-hand rule, with norm v w = v w sin(θ). V x W O W V Remark: Cross product of two vectors is another vector; which is perpendicular to the original vectors. W x V

3 Geometric definition of cross product v w is the area of the parallelogram formed by vectors v and w. Proof. V 0 V sin(0) The area A of the parallelogram formed by v and w is A = w ( v sin(θ) ) = v w. Definition Two vectors are parallel iff the angle in between them is θ = 0. W v w The non-zero vectors v and w are parallel iff v w = 0. Geometric definition of cross product Recall: v w is the area of a parallelogram. The closer the vectors v, w are to be parallel, the smaller is the area of the parallelogram they form, hence the shorter is their cross product vector v w. V x W W 0 1 V x W W 0 2 V V

4 Geometric definition of cross product Compute all cross products involving the vectors i, j, and k. Solution: Recall: i = 1, 0, 0, j = 0, 1, 0, k = 0, 0, 1. z k x i j y i j = k, j k = i, k i = j, i i = 0, j j = 0, k k = 0, i k = j, j i = k, k j = i. Cross product and determinants (Sect. 12.4)

5 Properties of the cross product (a) v w = (w v ), (b) v v = 0; (c) (a v ) w = v (a w) = a (v w), (d) u (v + w) = u v + u w, (e) u (v w) (u v ) w, (skew-symmetric); (linear); (linear); (not associative). Proof. Part (a) results from the right-hand rule and (b) from part (a). Parts (b) and (c) are proven in a similar ways as the linear property of the dot product. Part (d) is proven by giving an example. Properties of the cross product Show that the cross product is not associative, that is, u (v w) (u v) w. Solution: We prove this statement giving an example. We now show that i (i k) (i i) k = 0. Indeed, i (i k) = i ( j ) = (i j ) = k i (i k) = k, (i i) k = 0 j = 0 (i i) k = 0. We conclude that i (i k) (i i) k = 0. Recall: The cross product of parallel vectors vanishes.

6 Cross product and determinants (Sect. 12.4) Cross product in vector components The cross product of vectors v = v 1, v 2, v 3 and w = w 1, w 2, w 3 is given by v w = (v 2 w 3 v 3 w 2 ), (v 3 w 1 v 1 w 3 ), (v 1 w 2 v 2 w 1 ). Proof: Use the cross product properties and recall the non-zero cross products i j = k, and j k = i, and k i = j. Express v = v 1 i + v 2 j + v 3 k and w = w 1 i + w 2 j + w 3 k, then v w = (v 1 i + v 2 j + v 3 k) (w 1 i + w 2 j + w 3 k). Use the linearity property. The only non-zero terms involve i j = k, and j k = i, and k i = j symmetric analogues. The result is and the v w = (v 2 w 3 v 3 w 2 ) i + (v 3 w 1 v 1 w 3 ) j + (v 1 w 2 v 2 w 1 ) k.

7 Cross product in vector components. Find v w for v = 1, 2, 0 and w = 3, 2, 1, Solution: We use the formula v w = (v 2 w 3 v 3 w 2 ), (v 3 w 1 v 1 w 3 ), (v 1 w 2 v 2 w 1 ) v w = [(2)(1) (0)(2)], [(0)(3) (1)(1)], [(1)(2) (2)(3)] v w = (2 0), ( 1), (2 6) v w = 2, 1, 4. Exercise: Find the angle between v and w above, using both the cross and the dot products. Verify that you get the same answer. Cross product and determinants (Sect. 12.4)

8 Determinants to compute cross products. Remark: Determinants help remember the v w components. Recall: (a) The determinant of a 2 2 matrix is given by a b c d = ad bc. (b) The determinant of a 3 3 matrix is given by a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c determinants are used to find 3 3 determinants. Determinants to compute cross products. The formula to compute determinants of 3 3 matrices can be used to find the the cross product v w, where v = v 1, v 2, v 3 and w = w 1, w 2, w 3, as follows i j k v w = v 1 v 2 v 3 w 1 w 2 w 3 Proof: Indeed, a straightforward computation shows that i j k v 1 v 2 v 3 w 1 w 2 w 3 = (v 2w 3 v 3 w 2 ) i (v 1 w 3 v 3 w 1 ) j +(v 1 w 2 v 2 w 1 ) k.

9 Determinants to compute cross products. Given the vectors v = 1, 2, 3 and w = 2, 3, 1, compute both w v and v w. Solution: We need to compute the following determinant: i j k w v = w 1 w 2 w 3 v 1 v 2 v 3 = i j k The result is w v = (9 2) i ( 6 1) j +( 4 3) k w v = 7, 7, 7. The properties of the determinant imply v w = w v. Hence, v w = 7, 7, 7. Cross product and determinants (Sect. 12.4)

10 Triple product and volumes Definition The triple product of the vectors u, v, w, is the scalar u (v w). Remarks: (a) The triple product of three vectors is a scalar. (b) The parentheses are important. First do the cross product, and only then dot the resulting vector with the first vector. (Cyclic rotation formula for triple product) u (v w) = w (u v ) = v (w u). Triple product and volumes The number u (v w) is the volume of the parallelepiped determined by the vectors u, v, w. W x V 0 U h V W A Proof: Recall the dot product: x y = x y cos(θ). Then, u (v w) = u v w cos(θ) = h v w. v w is the area A of the parallelogram formed by v and w. So, u (v w) = h A, which is the volume of the parallelepiped formed by u, v, w.

11 The triple product and volumes Compute the volume of the parallelepiped formed by the vectors u = 1, 2, 3, v = 3, 2, 1, w = 1, 2, 1. Solution: We use the formula V = u (v w). We must compute the cross product first: i j k v w = = (2 + 2) i (3 1) j + ( 6 2) k, that is, v w = 4, 2, 8. Now compute the dot product, u (v w) = 1, 2, 3 4, 2, 8 = , that is, u (v w) = 24. We conclude that V = 24. The triple product and volumes Remark: The triple product can be computed with a determinant. If u = u 1, u 2, u 3, v = v 1, v 2, v 3, and w = w 1, w 2, w 3, then u 1 u 2 u 3 u (v w) = v 1 v 2 v 3 w 1 w 2 w 3. Compute the volume of the parallelepiped formed by the vectors u = 1, 2, 3, v = 3, 2, 1, w = 1, 2, 1. Solution: u (v w) = = (1)(2 + 2) (2)(3 1) + (3)( 6 2), that is, u (v w) = = 24. Hence V = 24.

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