A FIRST COURSE IN LINEAR ALGEBRA An Open Text by Ken Kuttler R n : The Cross Product Lecture Notes by Karen Seyffarth Adapted by LYRYX SERVICE COURSE SOLUTION Attribution-NonCommercial-ShareAlike (CC BY-NC-SA) This license lets others remix, tweak, and build upon your work non-commercially, as long as they credit you and license their new creations under the identical terms. R n : The Cross Product Page /
The Cross Product Definition Let u = [ u u u 3 ] T and v = [ v v v 3 ] T. Then u v = u v 3 u 3 v (u v 3 u 3 v ) u v u v Note. u v is a vector that is orthogonal to both u and v. A helpful way to remember: i j k u v = u u u 3 v v v 3, where i =., j =, k =. R n : The Cross Product The Cross Product Page /
Computing the Cross Product Problem Find u v for u = Solution We will use the equation:, v = u v = 3. u v 3 u 3 v (u v 3 u 3 v ) u v u v Therefore, u v = ( )() ()( ) (()() ()(3)) ()( ) ( )(3) = 3 5 R n : The Cross Product The Cross Product Page 3/
Properties of the Cross Product Theorem Let u, v and w be in R 3. u v is a vector. u v is orthogonal to both u and v. 3 u = and u =. 4 u u =. 5 u v = ( v u). 6 (k u) v = k( u v) = u (k v) for any scalar k. 7 u ( v + w) = u v + u w. 8 ( v + w) u = v u + w u. R n : The Cross Product The Cross Product Page 4/
Problem Find all vectors orthogonal to both u = [ 3 ] T and v = [ ] T. Solution u v = i j 3 k = 5 i + j k = 5. Any scalar multiple of u v is also orthogonal to both u and v, so 5 t, t R, gives all vectors orthogonal to both u and v. R n : The Cross Product The Cross Product Page 5/
i.e., x + y = 4 is an equation of the plane. R n : The Cross Product The Cross Product Page 6/ Problem Let A = (,, ), B = (,, ) and C = (,, 3) be point in R 3. These point do no all lie on the same line (how can you tell?). Find an equation for the plane containing A, B, and C. Solution A n B C AB and AC lie in the plane, so n = AB AC is a normal to the plane. AB = 3 Therefore, the plane has equation x + y =, AC = 3 = 4,, and n =.
Distance between Skew Lines Problem Given two lines x L : y = z 3 + s and L : x y z = + t, A. Find the shortest distance between L and L. B. Find the shortest distance between L and L, and find the points P on L and Q on L that are closest together. R n : The Cross Product The Cross Product Page 7/
Solution A Solution P P Q Choose P (3,, ) on L and P (,, ) on L. P Let d = and d = denote direction vectors for L and L, respectively. R n : The Cross Product The Cross Product Page 8/
Solution (continued) P (,, ) Q P P (3,, ) d =, d = The shortest distance between L and L is the length of the projection of P P onto n = d d. P P = and n = = 3 ( ) P P n proj n P P = n n, and proj n P P = P P n. n Therefore, the shortest distance between L and L is 8 4 = 4 7 4. R n : The Cross Product The Cross Product Page 9/
Solution B Solution P (,, ) Q P P (3,, ) d = P = 3 + s + s s, d = ; for some s R; Q = + t for some t R. t Now PQ = [ s + t s + s + t ] T is orthogonal to both L and L, so PQ d = and PQ d =, i.e., 3s t = s + 5t =. R n : The Cross Product The Cross Product Page /
Solution B Solution (continued) This system has unique solution s = 5 7 and t = 7. Therefore, ( 6 P = P 7, ) ( 8 7, and Q = Q 7 7,, ). 7 The shortest distance between L and L is PQ. Since P = P ( 6 7, ) 7, 7 and Q = Q ( 8 7,, ), 7 PQ = 7 8 4 7 6 = 7 8 4, R n : The Cross Product The Cross Product Page /
Solution B Solution (continued) Therefore PQ = 7 8 4 PQ = 4 4 = 4. 7 7 The shortest distance between L and L is 4 7 4. R n : The Cross Product The Cross Product Page /
Area and Volume The Lagrange Identity If u, v R 3, then Proof. Write u = x y z u v = u v ( u v). and v = x y z, and work out all the terms. R n : The Cross Product The Cross Product Page 3/
The length of the cross product As a consequence of the Lagrange Identity and the fact that u v = u v cos θ, we have u v = u v ( u v) = u v u v cos θ = u v ( cos θ) = u v sin θ. Taking square roots on both sides yields, u v = u v sin θ. Note that since θ π, sin θ. If θ = or θ = π, then sin θ =, and u v =. This is consistent with our earlier observation that if u and v are parallel, then u v =. R n : The Cross Product The Cross Product Page 4/
Area of a Parallelogram Theorem Let u and v be nonzero vectors in R 3 with included angle θ. u v = u v sin θ, and is the area of the parallelogram defined by u and v. u and v are parallel if and only if u v =. Proof of area of parallelogram. The area of the parallelogram defined by u and v is u h, where h is the height of the parallelogram. sin θ = θ v u h h v, implying that h = v sin θ. Therefore, the area is u v sin θ. R n : The Cross Product The Cross Product Page 5/
Area of a Triangle Problem Find the area of the triangle having vertices A(3,, ), B(,, ) and C(,, ). Solution The area of the triangle is half the area of the parallelogram defined by AB and AC. AB = and AC = 3 3. Therefore AB AC =, so the area of the triangle is AB AC =. R n : The Cross Product The Cross Product Page 6/
The Box Product Let u = u u u 3, v = v v v 3, and w = w w w 3. Then u v w 3 v 3 w u ( v w) = u u 3 (v w 3 v 3 w ) v w v w = u (v w 3 v 3 w ) u (v w 3 v 3 w ) + u 3 (v w v w ) v = u w v 3 w 3 u v w v 3 w 3 + u 3 v w v w u v w = u v w u 3 v 3 w 3. R n : The Cross Product The Box Product Page 7/
The Box Product Theorem u If u = u, v = u 3 v v v 3, and w = u ( v w) = det w w w 3 Shorthand: u ( v w) = det [ u v w ].. Then the box product is u v w u v w u 3 v 3 w 3 Theorem The order of the dot and cross product operations in computing the box product can be interchanged: ( u v) w = u ( v w).. R n : The Cross Product The Box Product Page 8/
The Volume of a Parallelepiped Theorem The volume of the parallelepiped determined by the three vectors u, v, and w in R 3 is u ( v w). Problem Findthe volume ofthe parallelepiped determined by the vectors u =, v =, and w =. R n : The Cross Product The Box Product Page 9/
Solution The volume of the parallelepiped is u ( v w). Since u ( v w) = det [ u v w ], and det =, the volume of the parallelepiped is = cubic units. R n : The Cross Product The Box Product Page /