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Math 3335. the Dot and Cross Products Throughout This Course I

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Math 3335. the Dot and Cross Products Throughout This Course I Math 3335. The Dot and Cross Products Throughout this course I will use the notation e1, e2, e3 to represent the so-called { } standard basis of R3 . Specifically 1 0 0 x1 e1 = 0 , e2 = 1 , e3 = 0 x = x2 = x1 e1 + x2 e2 + x3 e3. ⇒ 0 0 1 x3 Let x and y be any two vectors from R3 . The dot product of x and y is denoted by x y, and it is scalar valued; i.e. x y R. The cross product of x and y is denoted by · · ∈ x y, and it is vector valued; i.e. x y R3 . × × ∈ Both the dot and cross product are bilinear. That is, for any scalars α and β and any vector z we have (αx + βy) z = α (x z)+ β (y z), · · · z (αx + βy)= α (z x)+ β (z y), · · · (αx + βy) z = α (x z)+ β (y z), × × × z (αx + βy)= α (z x)+ β (z y). × × × Therefore 3 3 3 3 x y = xiyj (ei ej) and x y = xiyj (ei ej). · X X · × X X × i=1 j=1 i=1 j=1 This tells us that the dot product is uniquely defined by nine scalars, (ei ej), and the · cross product is uniquely defined by nine vectors, (ei ej). × A fundamental physical principle states that laws of physics should never depend on the particular orientation of the observer. Mathematically this principle is equivalent to the concept of rotational invariance. It is therefore natural to ask that both the dot product and cross product be rotationally invariant. Surprisingly, this added requirement will essentially determine both uniquely. 1. A proper rotation matrix R satisfies RT R = I and has det(R) = +1 so that it preserves orientation. Consider the following three matrices. 1 0 0 cos θ 0 sin θ cos θ sin θ 0 R1 = 0 cos θ sin θ , R2 = 0 1 0 , R3 = sin θ cos θ 0 . − 0 sin θ cos θ sin θ 0 cos θ 0 0 1 − − (a) Verify each R above is a proper rotation. (b) Determine the eigenvalues for each. (c) Investigate the action of each when applied to e1 , e2 and e3 . 2. Prove that the product of any two rotation matrices is a rotation matrix. Rotation invariance for the dot product says (Rx Ry)=(x y) · · for all vectors x and y and any rotation matrix R. Now what can we conclude from this? ◦ Let R be a 90 rotation which takes e1 to e2 to see (e1 e1)=(Re1 Re1)=(e2 e2). · · · Similarly see that (e1 e1)=(e3 e3) and so · · (e1 e1)=(e2 e2)=(e3 e3). · · · ◦ Next, let R be a 180 rotation which leaves e1 fixed and takes e2 to e2 . This gives − (e1 e2)=(Re1 Re2)=(e1 e2)= (e1 e2), which implies we must have (e1 e2)=0. · · ·− − · · Similar 180◦ rotations yield (ei ej)=0 for every i = j . · 6 Therefore, up to a free multiplicative constant, e.g. (e1 e1), a bilinear and rotationally · invariant scalar product necessarily takes the form 3 3 (1) x y == xiyj (ei ej) · X X · i=1 j=1 = x1y1 + x2y2 + x3y3. The right hand side of formula (1) gives what is called the dot product. 3. Let the dot product x y be given by formula (1) above. Its form was deduced as · a necessary consequence of rotation invariance. Here you will show it is in fact invariant under any rotation. (a) Regard vectors x and y as 3 1 column matrices. Verify that × x y = xT y. (b) Use this to see Rx Ry = xT RT Ry and then conclude the dot product · · is invariant under any rotation. (c) The Euclidean norm of a vector x R3 is given by ∈ x = √x x. Prove that Rx = x for any rotation R. || || · || || || || 4. Suppose x = x1e1 with x1 > 0 and y = y1e1 + y2e2 with y = 0. (a) By plane 6 trigonometry conclude that x y cos θ = x y where θ is the angle between x and y. || |||| || · (b) Use rotational invariance to conclude this formula is valid for any two nonzero three dimensional vectors x and y. Deducing the form of the cross product is a bit more involved compared to what was done for the dot product. We again ask that the rotation invariance property be necessarily satisfied, but here since the cross product is vector valued, this property requires (Rx Ry)= R (x y) × × for any proper rotation matrix R. First, for any i we must have (ei ei) = 0. To see × this, let R have ei as its axis of rotation (i.e. Rei = ei ) to see that since (ei ei)=(Rei Rei)= R(ei ei) × × × we must have (ei ei)= αei for some scalar α. (For this particular R we have Rx = x × x = αei since the eigenspace associated to eigenvalue λ = 1 is spanned by ei .) Now ⇔ ◦ use this and rotate by 180 with rotation axis perpendicular to ei to see αei =(ei ei)=( ei ei)= αei αei = 0. × − × − − ⇒ ◦ To deduce the form of the off-diagonal terms consider first (e1 e2) and a 180 rotation × R about e3 . This gives (e1 e2)=( e1 e2)=(Re1 Re2)= R(e1 e2) (e1 e2)= βe3 × − × − × × ⇒ × for some scalar β . This and a 90◦ rotation R also gives (e2 e1)=(Re1 Re2)= (Re1 Re2)= βRe3 = βe3. × × − − × − − ◦ Rotate these two relations by 90 about e1 and then again about e2 to finally get (e1 e2)= (e2 e1)= βe3, × − × (e1 e3)= (e3 e1)= βe2, × − × − (e2 e3)= (e3 e2)= βe1. × − × Therefore, up to a free multiplicative constant, e.g. β above, a bilinear and rotationally invariant vector product necessarily takes the form 3 3 (2) x y = xiyj (ei ej) × X X × i=1 j=1 =(x2y3 x3y2) e1 (x1y3 x3y1) e2 +(x1y2 x2y1) e3. − − − − The right hand side of formula (2) gives what is called the cross product. 5. A cofactor expansion easily establishes the formula x1 y1 z1 det x2 y2 z2 =(x y) z. × · x3 y3 z3 (a) Use properties of the determinant to prove (x y) z =(z x) y =(y z) x. × · × · × · (b) Also prove (Mx My) Mz = det(M) ((x y) z) for any 3 3 matrix M . × · × · × (c) Prove the cross product is rotationally invariant. That is, (Rx Ry)= R(x y) for × × any proper rotation matrix R. Hint: Use part (b). 6. We showed above that (ei ej)= (ej ei) for every i and j . (a) Use this to prove × − × (x y) = (y x) for any two vectors x and y. (b) Use this to conclude x x = 0 × − × × for any vector x. (c) Prove that x y is perpendicular to both x and y. Hint: Use 5(a) × and part (b) of this exercise. For the two problems below, define 2 1 2 3 1 a = 3 , b = 1 , c = 1 , d = 2 , e = 0 . 5 1 1 1 1 − 7. Compute the following. (a) a b (b) c d (c) (a b) e (d) (c d) e × × × · × · 8. Compute the following. (a) (a b) c (b) a (b c) (c)(c d) e (d) c (d e) × × × × × × × ×.
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