2. Let u = (ui, u2, us) andu = (vi,V2, ts) be vectors in R3. The cross product o

Question

2. Let u = (ui, u2, us) andu = (vi,V2, ts) be vectors in R3. The cross product of u and v is defined by A convenient way to remember this formula is through the following symbolic ex- pression: where i = ( 1 , 0, 0 ), j = (0,1,0), k = (0, 0, 1). (i) Strictly speaking, the above expression does not make sense since the determi- nant is an operation which transforms a real n x n matrix whose into a single real number. Even so, if one ignores the fact that i, j, and k are actually vectors and computes the determinant of the above matrix in the usual way, the result is exactly the formula for u x v. Verify this fact.

Explanation / Answer

(i) Let A =

i

j

k

u1

u2

u3

v1

v2

v3

The cofactor of i is u2v3- u3v2, the cofactor of j is u3v1- u1v3 and the cofactor of k is u1v2- u2v1. Therefore, det(A) = i(u2v3- u3v2)+j(u3v1- u1v3)+k(u1v2- u2v1).

(ii) We have vxu =det(B), where B =

i

j

k

v1

v2

v3

u1

u2

u3

Thus vxu = i(v2u3-v3u2)+k(v3u1 –v1u3)+k(v1u2-v2 u1) = -[ i(u2v3- u3v2)+j(u3v1- u1v3)+k(u1v2- u2v1)] = -uxv.

(iii) We have u.(uxv) = (u1,u2,u3).( u2v3- u3v2, u3v1- u1v3 , u1v2- u2v1) = u1(u2v3- u3v2)+u2(u3v1- u1v3) + u3(u1v2- u2v1)= u1u2v3-u1u3v2+ u2u3v1-u1u2v3+ u1u3v2-u2u3v1 = 0.

(iv) We have u.(uxv) = 0 (as per part (iii) above).On interchanging u and v, we get v.(vxu) = 0.Now, as per part(ii) above, we have vxu = -(uxv) so that v.[-(uxv)] = 0 or, - v.(uxv)= 0 so that v.(uxv)= 0.

(v) From part(iii)and (iv) above, we have u.(uxv) = 0 and v.(uxv)= 0 so that uxv is orthogonal to both u and v. Now, if v = cu for some real scalar c, then uxv = uxcu = i(cu2u3- cu3u2)+j(cu3u1- cu1u3)+k(cu1u2- cu2u1) = 0

i

j

k

u1

u2

u3

v1

v2

v3

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