This identity relates norms dot products and cross

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This identity relates norms, dot products, and cross products. In terms of the angle θ between x and y , we have from p. 1–17 the formula x y = k x k k y k cos θ . Thus, k x × y k = k x k k y k sin θ. This result completes the geometric description of the cross product, up to sign. The vector x × y is orthogonal to the plane determined by 0, x and y , and its norm is given by the formula we have just derived. An even more geometric way of saying this is that the norm of x × y is the area of the parallelogram with vertices 0, x , y , x + y : x y y x+y x θ
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6 Chapter 7 PROBLEM 7–3. The Lagrange identity can easily be generalized. Using the following outline, prove that for x , y , u , v R 3 , ( x × y ) ( u × v ) = ( x u )( y v ) - ( x v )( y u ) . Outline: the right side equals ˆ X i x i u i ! ˆ X j y j v j ! - ˆ X i x i v i ! ˆ X j y j u j ! = X i,j x i y j ( u i v j - u j v i ) (why?) = X i,j x j y i ( u j v i - u i v j ) (why?) = 1 2 X i,j ( x i y j - x j y i )( u i v j - u j v i ) (why?) = X i<j ( x i y j - x j y i )( u i v j - u j v i ) (why?) = ( x × y ) ( u × v ) . C. Triple products The formal representation of x × y as a determinant enables us to calculate inner products readily: for any u R 3 ( x × y ) u = det ˆ ı ˆ ˆ k x 1 x 2 x 3 y 1 y 2 y 3 u 1 ˆ ı + u 2 ˆ + u 3 ˆ k · = det u 1 u 2 u 3 x 1 x 2 x 3 y 1 y 2 y 3 . (Notice how clearly this formula displays the fact that ( x × y ) x = ( x × y ) y = 0.) This number ( x × y ) u is called the scalar triple product of the vectors x , y , and u . The formula we have just given makes it clear that
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Cross product 7 ( x × y ) u = ( u × x ) y = ( y × u ) x = - ( y × x ) u = - ( x × u ) y = - ( u × y ) x. Thus we have the interesting phenomenon that writing x , y , u in order gives ( x × y ) u = x ( y × u ) . So the triple product doesn’t care which you call × and which you call . For this reason, it’s frequently written [ x, y, u ] = ( x × y ) u. PROBLEM 7–4. The vector triple product is ( x × y ) × u . It can be related to dot products by the identity ( x × y ) × u = ( x u ) y - ( y u ) x. Prove this by using Problem 7–3 to calculate the dot product of each side of the proposed formula with an arbitrary v R 3 . PROBLEM 7–5. Prove quickly that the other vector triple product satisfies x × ( y × u ) = ( x u ) y - ( x y ) u. The identities in Problems 7–4 and 7–5 can be remembered by expressing the right side of each as vector triple product = (outer far) near - (outer near) far . The scalar triple product is exactly the device that enables us to compute volumes of parallelepipeds in R 3 . Consider three vectors x , y , u in R 3 . They generate a region D called a parallelepiped as follows: D = { ax + by + cu | 0 a 1 , 0 b 1 , 0 c 1 } . PROBLEM 7–6. Prove that [ x × y, y × u, u × x ] = [ x, y, u ] 2 .
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8 Chapter 7 O y u x We use the usual intuitive definition of volume as the area of the base times the altitude. If we regard x and y as spanning the parallelogram which is the base, its area is of course k x × y k .
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