The altitude is then calculated as the norm of the

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The altitude is then calculated as the norm of the projection of the side u onto the direction orthogonal to the base; of course, x × y has the required direction: θ x y O plane of 0, x, y u In terms of the angle θ between x × y and u , the altitude is

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Cross product 9 k u k | cos θ | = k u k | ( x × y ) u | k x × y k k u k = | ( x × y ) u | k x × y k . Thus the volume of D is vol( D ) = | ( x × y ) u | . It is therefore computed as the absolute value of the scalar triple product of the three spanning edges of the parallelepiped. In terms of the determinant, vol( D ) = fl fl fl fl fl det u 1 u 2 u 3 x 1 x 2 x 3 y 1 y 2 y 3 fl fl fl fl fl . Notice in particular that vol( D ) = 0 ⇐⇒ the vectors x , y , u are linearly dependent ⇐⇒ the parallelepiped D lies in a two dimensional subspace of R n . D. Orientation We now have the cross product characterized geometrically, up to a sign. Given linearly independent x , y R 3 , the cross product has direction orthogonal to both x and y , and norm equal to k x k k y k sin θ , where θ is the angle between the vectors x and y . There are exactly two vectors with these properties, and we now want to discuss a geometric way to single out the cross product. The concept we need is important in general dimensions, so we shall deal with R n for any n . DEFINITION. A frame for R n is an ordered basis. That is, we have n linearly independent vectors ϕ 1 , . . . , ϕ n , and they are written in a particular order. For example, { ˆ ı, ˆ , ˆ k } , { ˆ ı + 2ˆ , ˆ , - ˆ k } , { ˆ ı, ˆ k, ˆ } are three distinct frames for R 3 . DEFINITION. Let { ϕ 1 , . . . , ϕ n } and { ψ 1 , . . . , ψ n } be frames for R n . Then there exists a unique n × n matrix A such that ϕ i = i , 1 i n.
10 Chapter 7 This matrix is necessarily invertible. We say the two frames have the same orientation if det A > 0; and opposite orientation if det A < 0. PROBLEM 7–7. In the above definition we have used the column vector representa- tion of vectors. We may then write the corresponding n × n matrices Φ = ( ϕ 1 . . . ϕ n ) , Ψ = ( ψ 1 . . . ψ n ) . Prove that the frames have the same orientation ⇐⇒ det Φ and det Ψ have the same sign. PROBLEM 7–8. Prove that “having the same orientation” is an equivalence relation for frames. (Thus, as far as orientation is concerned, there are just two types of frames.) PROBLEM 7–9. Prove that if { ϕ 1 , . . . , ϕ n } and { ψ 1 , . . . , ψ n } happen to be orthonor- mal frames, the matrix A in the definition is in O ( n ); and the frames have the same orientation ⇐⇒ A SO( n ). Let us say that the coordinate frame { ˆ e 1 , . . . , ˆ e n } (also written { ˆ ı, ˆ , ˆ k } in the case of R 3 ) has the standard orientation. The corresponding matrix is the identity matrix I , and det I = 1. Thus we say that an arbitrary frame { ϕ 1 , . . . , ϕ n } has the standard orientation if the corresponding matrix Φ = ( ϕ 1 . . . ϕ n ) has det Φ > 0.

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