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8) and collecting up like terms we obtain z1 z2 = a1 a2 − b1 b2 − c1 c2 − d1 d2 + i(a1 b2 + b1 a2 ) + j(a1 c2 + c1 a2 ) + k(a1 d2 + d1 a2 ) + ijb1 c2 + ikb1 d2 + jic1 b2 + jkc1 d2 + kid1 b2 + kjd1 c2 . 9) But this, too, has some undefined terms: ij, ik, ji, jk, ki, kj. 10) which when substituted into Eq. 9) produces z1 z2 = a1 a2 − b1 b2 − c1 c2 − d1 d2 + i(a1 b2 + b1 a2 ) + j(a1 c2 + c1 a2 ) + k(a1 d2 + d1 a2 ) + kb1 c2 − jb1 d2 − kc1 b2 + ic1 d2 + jd1 b2 − id1 c2 . 11) Collecting up like terms we obtain z1 z2 = a1 a2 − (b1 b2 + c1 c2 + d1 d2 ) + i(a1 b2 + b1 a2 + c1 d2 − d1 c2 ) + j(a1 c2 + c1 a2 + d1 b2 − b1 d2 ) + k(a1 d2 + d1 a2 + b1 c2 − c1 b2 ).

96) As an example, let’s find the vector orthogonal to v1 × v2 where v1 = 1 v3 = 0 1 v3 = k which is correct. 0 0 T and v2 = 0 0 0 1 1 i+ j+ 0 0 0 0 1 0 k 1 0 T . 99) 36 Geometric algebra for computer graphics The vector product is also known as the cross product to reflect the ‘×’ symbol used as the operator. Commutative Law a × b = b × a The commutative law does not hold for the vector product. Given i a × b = xa xb = ya yb j ya yb k za zb z za i+ a zb zb x xa j+ a xb xb ya k yb a × b = (ya zb − yb za )i + (za xb − zb xa )j + (xa yb − xb ya )k.

The complex plane, or Argand diagram, uses two orthogonal axes to locate a complex number: the horizontal axis is the real axis whilst the vertical axis is the imaginary axis. 2 illustrates this concept and shows how the complex number z = a + ib is viewed as a 2D point with coordinates (a, b) or as a vector z with components [a b]T . 2. 36) and allows Eq. 10) to be written as zz ∗ = |z|2 . 37) The polar angle θ between the real axis and the position vector is identified using the inversetangent function (tan−1 ).

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