from IPython.display import YouTubeVideo
YouTubeVideo("zBlAGGzup48")
The Dot Product
takes two vectors and returns a scalar: ${\bf{a}}, {\bf{b}} \longmapsto {\bf{a}}\cdot {\bf{b}}$.
Direction Cosines
Orthogonal
vectors${\bf{a}} \perp {\bf{b}}$, read ${\bf{a}}$ is orthogonal to ${\bf{b}}$, if ${\bf{a}} \cdot {\bf{b}} = 0$.
${\bf{a}} \cdot {\bf{b}} = |{\bf{a}}||{\bf{b}}| \cos (\theta).$
import numpy as np
a = np.array([4,7,3]); b=np.array([0,1,1])
np.inner(a, b)
10
np.cross(a,b)
array([ 4, -4, 4])
c = np.array([1,1,1])
np.inner(np.cross(a,b),c)
4
In $\mathbb{R}^2$, we have the equation for a line
:
$$ ax + by = c.$$
Of course, we can solve for $y$ to rewrite the equation in slope-intercept form
:
$$by = c - ax$$
$$y = -\frac{a}{b} x + \frac{c}{b}.$$
In $\mathbb{R}^3$, we can form a similar expression and this describes a plane
:
$$ a x + by + cz = d.$$
We can solve for $z$:
$$cz = d -ax - by$$
$$z = -\frac{a}{c} x - \frac{b}{c}y + \frac{d}{c}$$
Another natural way to express a plane
is to normalize so that $d=1$. You can do this by dividing through by $d$.
Natural Questions!