Stuth Iomadachaidh-tharsainn a’ cleachdadh Cho-phàirtean Vectar Aonaid
Is urrainn dhuinn an toradh-tarsainn obrachadh a-mach gu dìreach ma tha fios againn air co-phàirtean nam vectaran. Tha an dòigh-obrach mar an ceudna ris an toradh dotaig . An toiseach, bidh sinn ag iomadachadh nam vectaran aonad i , j , agus k . Tha an toradh vectar eadar na h-aon vectaran aonad neoni.
i x i = j x j = k x k = 0
Le bhith a’ toirt iomradh air a’ cho-aontar iomadachaidh vectar a chaidh a thoirt a-mach roimhe (A x B = AB pheacaidh θ) agus an togalach an-aghaidh co-mhalairteach de iomadachadh vectar (A x B = – B x A), an uairsin gheibh sinn:
i x j = -j x i = k
j x k = -k x j = i
k x i = – i x k = j
A-nis bidh sinn a’ cur an cèill vectaran A agus B a thaobh an co-phàirtean, a’ dì-sgaradh an toraidh agus a’ cleachdadh toradh vectaran aonaid.
A x B= (Axi + Ayj + Azk) x (Bxi + Byj + Bzk)
A x B = Axi x Bxi + Axi x Byj + Axi x Bzk +
Ayj x Bxi + Ayj x Byj + Ayj x Bzk +
Azk x Bxi + Azk x Byj + Azk x Bzk
A x B = AxBx (i x i) + AxBy (i x j) + Ax Bz (i x k) +
AyBx (j x i) + AyBy (j x j) + AyBz (j x k) +
AzBx (k x i) + AzBy (k x j) + AzBz (k x k)
Air sgàth i x i = j x j = k x k = 0 dan i x j = –j x i = k, j x k = –k x j = i, k x i = -i x k = j, mar sin:
A x B = AxBx (0) + AxBy (k+ Ax Bz (-j+
AyBx (-k+ AyBy (0) + AyBz (i+
AzBx (j+ AzBy (-i+ AzBz (0)
A x B = AxBy (k+ Ax Bz (-j+
Ay Bx ( -k ) + Ay Bz ( i ) +
A z B x ( j ) + A z B y ( - i )
A x B = AxBy (k+ Ax Bz (-j+ AyBx (-k+ AyBz (i+ AzBx (j+ AzBy (-i)
A x B = (AyBz - AzBy)i + (AzBx - Ax Bz)j + (AxBy - AyBx )k
Ma tha C = A x B , is iad seo na pàirtean de C :
Cx = Ay Bz – AzBy
Cy = A z B x – A x B z
Cz = A x B y – A y B x
