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On the two dimensional plane, we can define the perp product or outer product or exterior product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) by replacing \(\mathbf{a}\) with it’s perp vector\(\mathbf{a}^\perp\). The perp dot product is denoted by \(\perp\) and is given by

\[\mathbf{a}\perp\mathbf{b}=\mathbf{a}^\perp\cdot\mathbf{b} = a_xb_y-a_yb_x = \left|\begin{array}{rl}a_x&a_y\\b_x&b_y\end{array}\right|\]


\[\mathbf{a}\perp\mathbf{b} = |\mathbf{a}| |\mathbf{b}|\sin\theta\]

With \(\theta\) being the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\). When \(|\mathbf{a}|=|\mathbf{b}|=1\), then \(\sin\theta=\hat{\mathbf{a}}\perp\hat{\mathbf{b}}\)



Scalar Association



\[\mathbf{a}\perp\mathbf{b} =-(\mathbf{b}\perp\mathbf{a})\]

Additive Distribution

\[\mathbf{a}\perp(\mathbf{b}+\mathbf{c}) = (\mathbf{a}\perp\mathbf{b})+(\mathbf{a}\perp\mathbf{c})\]

Lagrange Identity


2D Cross Product

Calculating the cross product of two-dimensional vectors, embedded into the 3D space with z-components zero yields

\[\mathbf{a}\times\mathbf{b} = \left|\begin{array}{ccc}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}}\\ a_x & a_y & 0\\ b_x & b_y & 0\end{array}\right|= \left|\begin{array}{cc}a_x & a_y\\ b_x & b_y\end{array}\right|\hat{\mathbf{z}} = (a_xb_y - a_yb_x)\hat{\mathbf{z}} = (\mathbf{a}\perp\mathbf{b})\hat{\mathbf{z}}\]

With \(\hat{\mathbf{x}}, \hat{\mathbf{y}}\) and \(\hat{\mathbf{z}}\) being the orthonormal basis, the result suggests that the perp-product is the 2D version of the 3D cross-product. While the cross product gives a normal to two vectors, the perp-product gives a normal to one vector.

The fundamental property of the cross-product and perp-product support this finding:

\[\begin{array}{rll} \mathbf{a}\perp\mathbf{b} &= |\mathbf{a}| |\mathbf{b}|\sin\theta&\text{ (2D case)}\\ |\mathbf{a}\times\mathbf{b}|&=|\mathbf{a}||\mathbf{b}|\sin\theta&\text{ (3D case)} \end{array}\]

Which stems from the Lagrange identity:

\[\begin{array}{rll} (\mathbf{a}\perp\mathbf{b})^2+(\mathbf{a}\cdot\mathbf{b})^2&=|\mathbf{a}|^2|\mathbf{b}|^2&\text{ (2D case)}\\ |\mathbf{a}\times\mathbf{b}|^2 + (\mathbf{a}\cdot\mathbf{b})^2& = |\mathbf{a}|^2|\mathbf{b}|^2&\text{ (3D case)}\\ \end{array}\]


Area of a parallelogram

To compute the (signed) area of a parallelogram spanned by the vectors \(\mathbf{a}\) and \(\mathbf{b}\), the perp product can be used, as one interpretation is exactly the area of the parallelogram:

\[\begin{array}{rl} \text{area}(\text{▱}\mathbf{a}\mathbf{b}) &= bh\\ &= |\mathbf{a}||\mathbf{b}|\sin\theta \\ &= \mathbf{a}\perp\mathbf{b} \end{array}\]

Area of a triangle

To compute the (signed) area of a triangle given its vertices \(P_1, P_2\) and \(P_3\), we can form the vectors \(\mathbf{a}=P_2-P_1\) and \(\mathbf{b}=P_3-P_1\). Since a triangle is half of a parallelogram, we get the area of the triangle \(\Delta P_1P_2P_3\) as

\[\text{area}(\Delta P_1P_2P_3) = \frac{1}{2}(\mathbf{a}\perp\mathbf{b})\]

Which is a much more compact formula over the calculation using the determinant:

\[\text{area}(\Delta P_1P_2P_3) = \frac{1}{2}\left|\begin{array}{ccc}P_1^x & P_1^y & 1\\ P_2^x & P_2^y & 1\\ P_3^x & P_3^y & 1\\\end{array}\right|\]

The signed area of the solution is positive when the vertices \(P_1, P_2\) and \(P_3\) are oriented counterclockwise and negative when oriented clockwise. This way the sign can be used for testing the orientation of a triangle.

Is point on left or right side of a line?

The perp product can be used to test if a point \(P_3\) is on the left or right of a line formed by the points \(P_1\) and \(P_2\). The area is positive when the point is on the left of line \(P_1P_2\) and on the right when the area is negative. The area is 0 when the point is on the line.

\(\mathbf{a}=P_2-P_1\) and \(\mathbf{b}=P_3-P_1\) gives

\[\begin{array}{rl} \mathbf{a}\perp\mathbf{b}=0&\Leftrightarrow\mathbf{a}\text{ and }\mathbf{b}\text{ are collinear, i.e } |\theta|=\text{0° or 180°}\\ \mathbf{a}\perp\mathbf{b}>0&\Leftrightarrow P_3\text{ is on left of } P_1P_2\text{, i.e. }0<\theta<180^\circ\\ \mathbf{a}\perp\mathbf{b}<0&\Leftrightarrow P_3\text{ is on right of } P_1P_2\text{, i.e. }-180^\circ<\theta<0\\ \end{array}\]

Since the three points form a triangle, \(|\theta|\leq 180^\circ\).

Shortest direction from one vector to the other

The perp product indicates the shortest rotation to get from \(\mathbf{a}\) to \(\mathbf{b}\). That is

\[\mathbf{a}\perp\mathbf{b}>0 \Leftrightarrow \text{ direction is counter-clockwise}\]