Vectors in R2 and R3

$\gdef \N{\mathbb{N}} \gdef \Z{\mathbb{Z}} \gdef \Q{\mathbb{Q}} \gdef \R{\mathbb{R}} \gdef \C{\mathbb{C}} \gdef \setcomp#1{\overline{#1}} \gdef \sseq{\subseteq} \gdef \pset#1{\mathcal{P}(#1)} \gdef \covariant{\operatorname{Cov}} \gdef \of{\circ} \gdef \p{^{\prime}} \gdef \pp{^{\prime\prime}} \gdef \ppp{^{\prime\prime\prime}} \gdef \pn#1{^{\prime\times{#1}}} $

Operations¶

1. Addition
2. Subtraction
3. Scalar Multiplication

Length of \(\vec v\):¶

\(\vec v = (x, y,\dots)\Rightarrow \|\vec v\|=\sqrt{x^2+y^2+\dots}\)
\(\|c\cdot\vec v\|=|c|\cdot\|\vec v\|\)

Dot Product¶

In \(\R^n\):
let \(\vec v=(v_1,v_2,\dots,v_n)\)
let \(\vec u=(u_1,u_2,\dots,u_n)\)
\(\vec v\cdot\vec u = ((v_1\cdot u_1) + (v_2\cdot u_2)+\dots+(v_n\cdot u_n))\)

For any \(\vec v\): \(\vec v\cdot \vec v=x^2+y^2=\|\vec v\|^2\)
\(\|\vec v\|=\sqrt{\vec v\cdot\vec v}\)

Properties¶

If \(\vec u, \vec v, \vec w\) are vectors in \(\R^2\) or \(\R^3\) and \(k\) is a scalar, then:

• \(\vec u\cdot\vec v=\vec v\cdot\vec u\)
• \(\vec u\cdot(\vec v+\vec w)=\vec u\cdot\vec v+\vec u\cdot\vec w\)
• \(k(\vec u\cdot\vec v)=(k\vec u)\cdot\vec v=\vec u\cdot(k\vec v)\)
• \(\vec v\cdot\vec v\gt0\text{ if }\vec v\not=\vec0\text{ and }\vec v\cdot\vec v=0\text{ if }\vec v=\vec0\)

Angle between vectors¶

The angle \(\theta\) between u and v is defined as the angle ≤ 180° formed by u and v when their tails coincide.
The angle is calculated as follows: \(\(cos(\theta) = \frac{\vec u\cdot\vec v}{\|\vec u\|\cdot\|\vec v\|}\)\)
Example:
Proof is at 23:04
What if \(\theta\) is 0°? well, \(cos\theta = 1\) , so there ya go. If \(\theta\) is 180°, \(cos\theta=-1\), and there ya go too.
Conclusion: vectors are ⟂ \(\iff\) their dot product is zero. These are called orthogonal vectors.

Cross Product¶

• Only defined in \(\R^3\)
• Output is a vector, not a scalar

• Definition:
  Let \(\vec u\) = \(<x_1, y_1, z_1>\), \(\vec v\) = \(<x_2, y_2, z_2>\). Then,

  \[\vec u\times\vec v = <(y_1\cdot z_2 - y_2\cdot z_1), (x_2\cdot z_1 - z_1,\cdot z_2), (x_1\cdot y_2-x_2\cdot y_1)>\]

  A faster way to calculate it, (and the “normal way”) is as follows.

  1. Write out \(\vec u\) above \(\vec v\) in a \(2\times3\) matrix:

     \[\begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{bmatrix}\]

  2. (for k = 1, 2, 3): Calculate the determinant of 2x2 matrix remaining by ignoring \(k^{th}\) column, this becomes the \(k^{th}\) component of the cross product

  3. The middle component \((k=2)\) of \(\vec u \times \vec v\) is the negative of determinant calculated in step 2. (No changes for \(k=1, k=3\))
     - Corollaries:
  4. If u, v have same or opposite directions (i.e., \(\vec v = c\cdot\vec u\mid c \in \R\) , then parallelogram they form is degenerate with area of 0, hence: If the length of the crossproduct is 0, the crossproduct is the zero vector)
  5. Area of Triangle ABC is half area of parallelogram formed by u and v, hence, \(area = \frac12\|\vec u\times\vec v\|\)

Representing Lines and Planes¶

• Lines in \(\R^2\):
  • 
    • A new way to obtain this equation:
      Let \((x, y)\) be any point on \(\ell\). Since \(\ell\) goes through origin, \((x, y)\) is also a direction vector of \(\ell\).
      • i.e. Direction of \(\ell\) = direction of vector \((x, y)\).
    • Using considerations of slope, the vector \((-2, 1)\) is orthogonal to any vector in direction of \(\ell\).
      • We call \((-2, 1)\) a normal vector to \(\ell\).
    • Hence, \((-2, 1)\cdot(x, y) = 0 \implies -2x+y=0\) (see diagram)
      • Meaning, \(\vec n\cdot(x, y) = 0 \space | \space \forall (x, y) \in \ell\)

• Planes in \(\R^3\):
  • The set of directions ⟂ to a given vector in \(\R^3\) is a plane.
    • Ex. If \((a, b, c) = (0, 0, 1)\) [z-axis], then “perpendicular directions” are \((x, y)\) - plane: \((x, y, 0)\)
    • \(\lbrace (0, 0, 1)\cdot(x,y,0)\mid x,y\in\R\rbrace\)
  • Conclusion: ax + by + cz = 0 describes a plane through origin, to which \ is a normal vector.
    • For any vector \((x, y, z)\) in this plane, \((a,b,c)\cdot(x,y,z) = 0\)
  • General planes in \(\R^3\):
    • Let plane \(P_1\) be a plane through the origin with a normal vector of (2, 1, 3).
    • Equation of \(P_1\) is \(2x+y+3z = 0\).
    • Let plane \(P_2\) be parallel to \(P_1\), just shifted so it goes through \((6, 1, -3)\).
    • What’s the equation of \(P_2\)?
      • Let \(\vec v\) be one direction vector on \(P_2\).
      • Let \((x, y, z)\) be any point in \(P_2\).
      • \(\vec v = (x-6, y-1, z+3)\). (just because it’s a direction)
      • \(\text{The normal vector to } P_1 = \text{the normal vector to }P_2\)
      • This is because \(P_2\) is parallel to \(P_1\).
      • Equation is: \((x-6, y-1, z+3)\cdot\vec n = 0 \implies \dots \implies 2x+y+3z=4\)

      • Note the similarities between equations of parallel planes
        \(P_1\): \(2x+y+3z = 0\)
        \(P_2\): \(2x+y+3z = 4\)

    • Conclusion: Any plane in \(\R^3\) has equation of \(ax+by+cz=d\), where \((a, b, c)\) is a normal vector to the plane.
      • Ex: Find equation of plane with normal of \((-1, 4, 2)\) going through the point (0, 1, -3)
        • Use the normal vector to get a general equation: \(-x+4y+2z=d\)
        • Then, plug in our point \((0, 1, -3)\) to find d:
          • \(-1(0)+4(1)+2(-3) = -2\)
        • Putting it all together, we get: \(-x+4y+2z=-2\)

• Lines in \(\R^3\)

  • A line in \(\R^3\) cannot be described using a single equation. Instead, we describe it parametrically - that is, we describe it as a set of points using a parameter (usually t).
  • 
  • If \(\ell\) goes through origin and \(\vec v\) is any point on \(\ell\) (in example, \(\vec v=(3,-2,4)\)), then \(\ell\) consists exactly of all scalar multiples of \(\vec v\). \(\Rightarrow\) \(\ell=\{t\cdot\vec v\mid t\in\R\}\). This is the parametric form.

  • Cool, but what about lines not through the origin?

    • If \(\vec v\) is a direction vector of \(\ell\), then every point on \(\ell\) is of form \(P+t\cdot\vec v\) for some \(t\in\R\).
    • 
    • Hence, \(\ell=\{(2,1,-4)+t\cdot\vec v\mid t\in\R\}\)
    • We find \(\vec v\) by subtracting 2 points on line:
    \[\begin{aligned} \vec v &= Q-P\\&= (5,-1,-5)\Rightarrow\\ \ell &=\{(2, 1, -4) + t\cdot(5,-1,-5)\mid t\in\R\}\\ &=\{(2+5t,1-t,-4-5t)\mid t\in\R\} \end{aligned}\]

  • In summary

    1. Line in \(\R^3\) through origin:
       • \(\ell=\{t\cdot\vec u\mid t\in\R\}\)
    2. Line in \(\R^3\) not through origin:
       - \(\ell=\{\vec u+t\cdot\vec v\mid t\in\R\}\)
       - where \(\vec u\) is any direction vector of \(\ell\).
       - Usually, \(\vec u\) is found by subtracting 2 points on \(\ell\)
  • Summary
  • Cartesian Representation (equation):
    • Line in \(\R^2\): \(ax + by = c\)
    • Plane in \(\R^3\): \(ax+by+c=d\)
      • normal vector \(\vec n\) is orthogonal to all direction vectors in plane
  • Parametric representation (set of points):
    • Line (in \(\R^n\mid n\geqslant 2\)): \(\ell=\{\vec u+t\cdot\vec v\mid t\in\R\}\)
      • where \(\vec u\) is any point on \(\ell\) and \(\vec v\) is a direction vector of \(\ell\).
      • Neither \(\vec u\text{ or }\vec u\) is unique (because any scalar multiple of a direction vector is also a direction vector)
      • If \(\ell\) goes through the origin, then \(\vec u=(0, 0, 0)\), and adding it has no effect. Therefore, it can be written as: \(\ell=\{t\cdot\vec u\mid t\in\R\}\)
  • Another way to represent a line in \(\R^3\):
  • Two non-parallel planes in \(\R^3\) intersect at a line.
  • Each point on this line must satisfy the equations of both planes.
  • 
  • In example in diagram, any (x, y, z) on \(\ell\) must be a solution of the system \(\left\{\begin{aligned}x+y+z&=2\\-x+2y-2z&=1\end{aligned}\right.\)
  • Conversely, any solution of the system is a point on \(\ell\).

Tirgulim [Incomplete]¶