Euclidean Geometry

Definition Inner Product

• $a, b \in \mathbb{R}^{n}$,

\[\langle a, b \rangle := a \cdot b := \sum_{i=1}^{n} a_{i}b_{i} .\]

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Definition Cross Product

• $a, b \in \mathbb{R}^{3}$

\[\begin{eqnarray} a \times b & := & \norm{a} \norm{b} \sin(\theta) n \nonumber \\ & = & \left( \begin{array}{c} a_{2}b_{3} - a_{3}b_{2} \\ a_{3}b_{1} - a_{1}b_{3} \\ a_{1}b_{2} - a_{2}b_{1} \end{array} \right) \nonumber \end{eqnarray}\]

where $n$ is a unit vector perpendicular to the plane containing $a$ and $b$.

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Definition Outer Product

• $a \in \mathbb{R}^{m}$,
• $b \in \mathbb{R}^{n}$,

\[\begin{eqnarray} a \times b & := & \left( \begin{array}{c} a_{1} \\ \vdots \\ a_{m} \end{array} \right) \times \left( \begin{array}{c} b_{1} \\ \vdots \\ b_{n} \end{array} \right) \nonumber \\ & = & \left( \begin{array}{c} a_{1} \\ \vdots \\ a_{m} \end{array} \right) \left( \begin{array}{c} b_{1} \\ \vdots \\ b_{n} \end{array} \right)^{T} \nonumber \\ & = & \left( \begin{array}{cccc} a_{1}b_{1} & a_{1}b_{2} \cdots & a_{1}b_{m} \\ \vdots & & & \vdots \\ a_{n}b_{1} & a_{n}b_{2} \cdots & a_{n}b_{m} \end{array} \right) \nonumber . \end{eqnarray}\]

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Definition Exterior Product

• $a, b \in \mathbb{R}^{n}$

\[\begin{eqnarray} a \wedge b & = & \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} b_{j} e_{i} \wedge e_{j} \nonumber \end{eqnarray}\]

where $e_{i} \wedge e_{j} = -e_{j} \wedge e_{i}$.

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Example 1

• $a \in \mathbb{R}^{2}$,
• $b \in \mathbb{R}^{2}$,
• $e_{1}, e_{2}$
  • basis

\[\begin{eqnarray} a \wedge b & = & a_{1}b_{1} e_{1}\wedge e_{1} + a_{1}b_{2} e_{1}\wedge e_{2} + a_{2}b_{1} e_{2}\wedge e_{1} + a_{2}b_{2} e_{2}\wedge e_{2} \nonumber \\ & = & a_{1}b_{2} e_{1}\wedge e_{2} - a_{2}b_{1} e_{1}\wedge e_{2} \nonumber \\ & = & (a_{1}b_{2} - a_{2}b_{1}) e_{1}\wedge e_{2} . \nonumber \end{eqnarray}\]

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Example 2

• $a \in \mathbb{R}^{3}$,
• $b \in \mathbb{R}^{3}$,
• $e_{1}, e_{2}, e_{3}$
  • basis

\[\begin{eqnarray} a \wedge b & = & a_{1}b_{1} e_{1}\wedge e_{1} + a_{1}b_{2} e_{1}\wedge e_{2} + a_{1}b_{3} e_{1}\wedge e_{3} + a_{2}b_{1} e_{2}\wedge e_{1} + a_{2}b_{2} e_{2}\wedge e_{2} + a_{2}b_{3} e_{2}\wedge e_{3} + a_{3}b_{1} e_{3}\wedge e_{1} + a_{3}b_{2} e_{3}\wedge e_{2} + a_{3}b_{3} e_{3}\wedge e_{3} \nonumber \\ & = & a_{1}b_{2} e_{1}\wedge e_{2} + a_{1}b_{3} e_{1}\wedge e_{3} + a_{2}b_{1} e_{2}\wedge e_{1} + a_{2}b_{3} e_{2}\wedge e_{3} + a_{3}b_{1} e_{3}\wedge e_{1} + a_{3}b_{2} e_{3}\wedge e_{2} \nonumber \\ & = & a_{1}b_{2} e_{1}\wedge e_{2} + a_{1}b_{3} e_{1}\wedge e_{3} - a_{2}b_{1} e_{1}\wedge e_{2} + a_{2}b_{3} e_{2}\wedge e_{3} - a_{3}b_{1} e_{1}\wedge e_{3} - a_{3}b_{2} e_{2}\wedge e_{3} \nonumber \\ & = & (a_{1}b_{2} - a_{2}b_{1}) e_{1}\wedge e_{2} + (a_{1}b_{3} - a_{3}b_{1}) e_{1}\wedge e_{3} + (a_{2}b_{3} - a_{3}b_{2}) e_{2}\wedge e_{3} . \nonumber \end{eqnarray}\]

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2-d vectors

• $a, b, c, d \in \mathbb{R}^{2}$,

(1) $a$ and $b$ are perpendicular if

\[\langle a, b \rangle = a_{1}b_{1} + a_{2}b_{2} = 0.\]

(2) $a$ and $b$ are parallel if

\[a \times b = a_{1}b_{2} - a_{2}b_{1} = 0 .\]

(4) $c$ is on a line which contains $a$ and $b$ if

\[(b - a) \times (c - a) = (b_{1} - a_{1})(c_{2} - a_{2}) - (b_{2} - a_{2})(c_{1} - a_{1}) = 0 .\]

(5) $c$ is on a line segment $b - a$ if

\[(b - a) \times (c - a) = (b_{1} - a_{1})(c_{2} - a_{2}) - (b_{2} - a_{2})(c_{1} - a_{1}) = 0\]

and

\[\langle a - c, b - c \rangle \le 0 .\]

(7) The distance between the line $b - a$ and a point $c$ is given by

\[\begin{eqnarray} \norm{c - a} |\sin(\theta)| & = & \frac{ (b - a) \times (c - a) }{ \norm{x} } \nonumber \end{eqnarray}\]

(8) The minimum distance between the line segment $b - a$ and a point $c$ is given by

\[\begin{cases} \frac{ (b - a) \times (c - a) }{ \norm{x} } & (\langle a - c, b - c \rangle \neq 0) \\ c - a & (\text{otherwise}) \end{cases}\]

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(9) Two line seguments $b -a$ and $d - c$ are interesected if

\[\begin{eqnarray} (b - a) \times (c - a) (b - a) \times (d - a) & < & 0, \nonumber \\ (d - c) \times (a - c) (d - c) \times (b - c) & < &0. \nonumber \end{eqnarray}\]

(10) The intersection of two line seguments $b - a$ and $d - c$ is given by

\[\begin{eqnarray} a + \frac{ (d - c) \times (c - a) }{ (d - c) \times (b - a) } (b - a) & = & a + \frac{ (d_{1} - c_{1}) (c_{2} - a_{2}) - (d_{2} - c_{2}) (c_{1} - a_{1}) }{ (d_{1} - c_{1}) (b_{2} - a_{2}) - (d_{2} - c_{2}) (b_{1} - a_{1}) } (b - a) . \end{eqnarray}\]

Note that the equation cannot detect the case where the both lines are in parallel. Indeed, if $(b - a)$ and $(d - c)$ are in parallel,

\[(b - a) \times (d - c) = 0 .\]

Translation

Let $f(x) := a x + b$. The line which shifts $f(x)$ $m$ in a direction parallel to $f(x)$. Let $g(x)$ be the line. $g(x)$ has the same slope so $g(x) = ax + c$. There is $n$ such that

\[ax + c = a(x - n) + b .\]

By rearranging the equation

\[n = \frac{ b - c }{ a } .\]

The similarilty of right triangles implies

\[m:(b - c) = l:n\]

That is

\[l = \frac{ m n }{ (b - c) } = \frac{ m }{ a } .\]

By Pythagorean theorem,

\[\begin{eqnarray} & & n^{2} = l^{2} + m^{2} \nonumber \\ & \Leftrightarrow & n^{2} = \left( \frac{ m }{ a } \right)^{2} + m^{2} \nonumber \\ & \Leftrightarrow & n^{2} = \frac{ m^{2} + a^{2}m^{2} }{ a^{2} } \nonumber \\ & \Leftrightarrow & n & = & \frac{ m\sqrt{1 + a^{2}} }{ a } \nonumber \end{eqnarray}\]

Hence

\[\begin{eqnarray} & & \frac{(b - c)}{a} = \frac{ m\sqrt{1 + a^{2}} }{ a } \nonumber \\ & \Leftrightarrow & c = b - m\sqrt{1 + a^{2}} \nonumber \end{eqnarray}\]

then

\[g(x) = ax + b - m\sqrt{1 + a^{2}} .\]

Reference

• https://en.wikipedia.org/wiki/Outer_product
• https://en.wikipedia.org/wiki/Cross_product
• https://en.wikipedia.org/wiki/Exterior_algebra