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text: '#Motivation\nwant to incorporate rotors like $e^{\\text{I}\\theta}$ and scalars $n$ in the same system\n\nalso want $|a|^2 + |b|^2 = |a+b|^2$ to be incorporated into the product\n\nwant to follow standard laws of products in algebra as much as possible - distribute over vector sum\n\nbut, we don't want it to be commutative, because we've seen that rotations in 3d are not commutative, so it would lose its power to describe geometric relations if we kept that (scalars themselves will always be commutable with everything)\n\nthere is a product that satisfies all of these. it was hard to find, carefully designed, and has a lot of built up mathematical insight going into it. we will try to understand why it is the way it is\n## Contraction rule\nkey idea - look at $(a+b)^2 = a^2 + ab + ba + b^2$ (remember $ab$ and $ba$ should different in general, so we can't combine them into $2ab$)\npretty close to pythagorean theorem!\n\nif we set $a^2 = |a|^2$, this will almost get us to the pythagorean theorem:\n$|a+b|^2 = |a|^2 + ab + ba + |b|^2$\n\nit's good that there's the $ab + ba$ in there, because the pythagorean theorem is only true for vectors at right angles\nso $ab + ba$ should be zero exactly when $a$ and $b$ are at right angles\n\nwhat about when $a$ and $b$ are in the same direction?\nthen $|a+b|^2 = (|a| + |b|)^2 = |a|^2 + 2|a||b| + |b|^2$\nso in this case, $ab + ba$ should be $2|a||b|$\n\nso for right angles $ab = - ba$\n\nfor parallel vectors makes sense to have $ab = ba = |a||b|$\n\nimportant consequence: $\\frac{1}{a}=\\frac{a}{|a|^2}$ - actually need to use $a^{-1}$ bc non-commutative\n\n## Rotations\n\nfrom rotor theory, for right angles, $ae^{\\text{I}\\pi/2} = b$ for same magnitude\nso $e^{\\text{I}\\pi/2} = \\frac{ab}{|a|^2}$, so $ab = |a|^2e^{\\text{I}\\pi/2}$, when $|a| = |b|$\n\nthen if $b$ is diff magnitude, $a|b|/|a|e^{\\text{I}\\pi/2} = b$, so $ab = |a||b|e^{\\text{I}\\pi/2}$.\n\nthis means $|b||a|=-e^{\\text{I}\\pi/2}$. \n\none more thing we have to set, we will identify $e^{\\text{I}\\pi/2}$ with the unit bivector $\\text{I}$ itself.\n\nso for perpendicular angles, $ab = |a||b|I$\n\nthis means $I^2 = -1$\n\nso what is $ab$ in general?\n\nwrite $a = a_xx+a_yy$, $b = b_xx+b_yy$\n$ab = (a_xx + a_yy)(b_xx + b_yy) = a_xb_xx^2 + a_yb_xyx + a_xb_yxy+a_yb_yy^2 = a_xb_x + a_yb_y - a_yb_xI + a_xb_yI$\n\nscalar part + bivector part\n\nthis gives us an alternate way of writing angles\n\n$e^{\\text{I}\\pi/4} = \\frac{1 + I}{\\sqrt{2}}$\n\ncall the scalar part the cosine, and the bivector part the sine\n\n$e^{\\text{I}\\theta} = \\cos(\\theta) + \\text{I}\\sin(\\theta)$\n\n(btw this is where all those trig identities you hated come from - they are much easier to work with in exponential form)\n\n\n## Multivectors\n\nrotor is sum of scalar and bivector\n\nin general - we can add together any type of $k$-vector\n\ncan think of the different parts as different aspects of the transformation\n\ndifferent parts are called grades\n\n\n',
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