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Consider the following theorem. If a, b, and c are vectors and c is a scalar, then we have the following properties of the cross product. 1. ax b = -bxa 2. (ca) x b = c(a x b) = ax (cb) 3. ax (b + c) = axb + axc 4. (a + b) x c = axc + bxc 5. a (b x c) = (axb) c 6. ax (bx c) = (a c)b (a b)c Prove the property a x b = -b x a of the given theorem. Let a = (a₁, a2, a 3) and b = (b₁,b₂, b3). Then we get the following. axb = = (-1) = -bxa.

Consider the following theorem. If a, b, and c are vectors and c is a scalar, then we have the following properties of the cross product. 1. ax b = -bxa 2. (ca) x b = c(a x b) = ax (cb) 3. ax (b + c) = axb + axc 4. (a + b) x c = axc + bxc 5. a (b x c) = (axb) c 6. ax (bx c) = (a c)b (a b)c Prove the property a x b = -b x a of the given theorem. Let a = (a₁, a2, a 3) and b = (b₁,b₂, b3). Then we get the following. axb = = (-1) = -bxa.

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter5: Inner Product Spaces
Section5.2: Inner Product Spaces
Problem 101E: Consider the vectors u=(6,2,4) and v=(1,2,0) from Example 10. Without using Theorem 5.9, show that...
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Consider the following theorem. If a, b, and c are vectors and c is a scalar, then we have the following properties of the cross product. 1. ax b = -bxa 2. (ca) x b = c(a x b) = ax (cb) 3. ax (b + c) = axb + ax c 4. (a + b) x c = axc + bxc 5. a (b x c) = (axb).c 6. ax (bx c) = (a c)b (a b)c Prove the property a x b = -b x a of the given theorem. Let a = (a₁, a2, a 3) and b = (b₁,b₂, b3). Then we get the following. axb= = (-1) = -bxa.
Transcribed Image Text:Consider the following theorem. If a, b, and c are vectors and c is a scalar, then we have the following properties of the cross product. 1. ax b = -bxa 2. (ca) x b = c(a x b) = ax (cb) 3. ax (b + c) = axb + ax c 4. (a + b) x c = axc + bxc 5. a (b x c) = (axb).c 6. ax (bx c) = (a c)b (a b)c Prove the property a x b = -b x a of the given theorem. Let a = (a₁, a2, a 3) and b = (b₁,b₂, b3). Then we get the following. axb= = (-1) = -bxa.
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