Processing math: 0%
Google
 
Web unafbapune.blogspot.com

Sunday, March 31, 2013

 

Vector dot and cross products

  1. , aka Cauchy-Schwartz Inequality
  2. \| \vec{x} + \vec{y} \| \le \|\vec{x}\| + \|\vec{y}\| , aka Triangular Inequality
  3. \vec{a} \cdot \vec{a} = \|\vec{a}\|^2 , which can be easily verified.
  4. \vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos{\theta} , where \theta is the angle between \vec{a} and \vec{b}. Can be verified based on (3).
  5. \displaystyle \vec{a} \times \vec{b} = \|\vec{a}\| \|\vec{b}\| \sin{\theta} \, n , where n is the unit vector perpendicular to the plane containing \vec{a} and \vec{b}. Note the cross product between two vectors is applicable only in \mathbb{R}^3. Based on (4), there is a nice proof at Khan Academy.
  6. \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} (\vec{a} \cdot \vec{c}) - \vec{c} (\vec{a} \cdot \vec{b}) . Very optional with nice proof at Khan Academy.
  7. How to determine the linear equation, Ax + By + Cz = D, of a plane in \mathbb{R}^3 if given a point on the plane, and a normal vector, which is the vector perpendicular to the plane ?
  8. Conversely, how to determine the normal vector if given the linear equation of a plane in \mathbb{R}^3 ? Turns out if the plan is Ax + By + Cz = D, the normal vector \vec{n} is simply Ai + Bj + Ck, where i, j, k are the unit vectors of the respective x, y, z coordinates.
  9. Given a plane, Ax + By + Cz = D, and a point vector x_0i + y_0j + z_0k outside of the plane, how to determine the shortest distance d from the point vector to the plane ? Turns out: \displaystyle d = \frac{Ax_0 + By_0 + Cz_0 - D}{\sqrt{A^2 + B^2 + C^2}}
Footnotes:

Comments: Post a Comment

<< Home

This page is powered by Blogger. Isn't yours?