Question #131498

Suppose u and v are vectors in 3–space where u = (u1, u2, u3) and v = (v1, v2, v3). Evaluate u×v×u and v × u × u.


1
Expert's answer
2020-09-02T17:44:03-0400

Submit u=(u1,u2,u3)u=(u_1,u_2,u_3), v=(v1,v2,v3)v=(v_1,v_2,v_3).


Calculate vector product of uu and vv:

u×v=ijku1u2u3v1v2v3=i(u2v3u3v2)+j(u1v3u3v1)+k(u1v2u2v1)==(u2v3u3v2,u1v3u3v1,u1v2u2v1).u\times v=\begin{vmatrix} \vec i & \vec j & \vec k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} =\vec i(u_2v_3-u_3v_2)+\vec j(u_1v_3-u_3v_1)+\vec k(u_1v_2-u_2v_1)=\\ {}=(u_2v_3-u_3v_2,\,u_1v_3-u_3v_1,\,u_1v_2-u_2v_1)..


And obtain:

u×v×u=ijku2v3u3v2u1v3u3v1u1v2u2v1u1u2u3=i(u1u3v3u32v1u1u2v2+u22v1)+j(u2u3v3u32v2u12v2+u1u2v1)+k(u22v3u2u3v2u12v3+u1u3v1)=(u1u3v3u32v1u1u2v2+u22v1,u2u3v3u32v2u12v2+u1u2v1,u22v3u2u3v2u12v3+u1u3v1).u\times v\times u=\begin{vmatrix} \vec i & \vec j & \vec k \\ u_2v_3-u_3v_2 & u_1v_3-u_3v_1 & u_1v_2-u_2v_1 \\ u_1 & u_2 & u_3 \end{vmatrix} =\vec i(u_1u_3v_3-u_3^2v_1-\\ {}-u_1u_2v_2+u_2^2v_1)+\vec j(u_2u_3v_3-u_3^2v_2-u_1^2v_2+u_1u_2v_1)+\vec k(u_2^2v_3-u_2u_3v_2-\\ {}-u_1^2v_3+u_1u_3v_1)=(u_1u_3v_3-u_3^2v_1-u_1u_2v_2+u_2^2v_1,\,u_2u_3v_3-u_3^2v_2-\\ {}-u_1^2v_2+u_1u_2v_1,\,u_2^2v_3-u_2u_3v_2-u_1^2v_3+u_1u_3v_1).


According to vector product properties u×v×u=u×(v×u)=(v×u)×uu\times v\times u=u\times (v\times u)=-(v\times u)\times u, so

v×u×u=u×v×u=(u1u3v3u32v1u1u2v2+u22v1,u2u3v3u32v2u12v2+u1u2v1,u22v3u2u3v2u12v3+u1u3v1)=(u1u3v3+u32v1+u1u2v2u22v1,u2u3v3+u32v2+u12v2u1u2v1,u22v3+u2u3v2+u12v3u1u3v1).v\times u\times u=-u\times v\times u=-(u_1u_3v_3-u_3^2v_1-u_1u_2v_2+u_2^2v_1,\,u_2u_3v_3-u_3^2v_2-\\ {}-u_1^2v_2+u_1u_2v_1,\,u_2^2v_3-u_2u_3v_2-u_1^2v_3+u_1u_3v_1)=(-u_1u_3v_3+u_3^2v_1+u_1u_2v_2-\\ {}-u_2^2v_1,\,-u_2u_3v_3+u_3^2v_2+u_1^2v_2-u_1u_2v_1,\,-u_2^2v_3+u_2u_3v_2+u_1^2v_3-u_1u_3v_1).


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