MAT223: 2_Vector, Lines, and Planes

Vectors, Lines, and Planes §

Row Reduction (Gauss-Jordan Elimination) §

Augmented matrix: $\left[\begin{array}{cccc}1& 2& 1& 0\\ 2& 3& 1& 1\\ 3& 1& 2& 0\end{array}\right]$ from the System: $\{\begin{array}{cc}& x+2y+z=0\\ & 2x+3y+z=1\\ & 3x+y+2z=0\end{array}$

• First phase: work down from top, from left from right

  • Top row stays the same (as the tool row) that contains the pivot column

  $\left[\begin{array}{cccc}1& 2& 1& 0\\ 0& -1& -1& 1\\ 0& -5& -1& 0\end{array}\right]$ $$\begin{gathered} Row2-2\cdot Row1 \\ Row3-3\cdot Row1 \end{gathered}$$

  $\left[\begin{array}{cccc}1& 2& 1& 0\\ 0& -1& -1& 1\\ 0& 0& 4& -5\end{array}\right]$ $$\begin{gathered} (-) \\ Row3-5\cdot Row2 \end{gathered}$$

• Pause to figure out if there is inconsistent cases

  • $[000-5]$ For the last row would give no solution

  • $[0000]$ Would give many solutions

  • Or else, there is a single solution

  $\left[\begin{array}{cccc}1& 2& 1& 0\\ 0& 1& 1& -1\\ 0& 0& 1& -5/4\end{array}\right]$ $$\begin{gathered} (-) \\ \frac{1}{4} \end{gathered}$$

• Second phase: work from bottom up, left to right (use last row as tool row)

  $\left[\begin{array}{cccc}1& 2& 0& 5/4\\ 0& 1& 0& 1/4\\ 0& 0& 1& -5/4\end{array}\right]$ $\begin{array}{c}Row1-Row3\\ Row2-Row3\\ \end{array}$

  $\left[\begin{array}{cccc}1& 0& 0& 3/4\\ 0& 1& 0& 1/4\\ 0& 0& 1& -5/4\end{array}\right]$ $\begin{array}{c}Row1-2\cdot Row2\\ \\ \end{array}$

  $\{\begin{array}{cc}& x=3/4\\ & y=1/4\\ & z=-5/4\end{array}$ As the solution of this system of equations

Vector Lines §

• Let $\ell$ be a line and let $\vec{d}$ and $\vec{p}$ be vectors (needs 2 distinct vectors)

  $$\ell =\{\vec{x}\mid \vec{x}=t\vec{d}+P\text{for some}t\in \mathbb{R}\}$$

• The vector equation:

  $$\vec{x}=t\vec{d}+\vec{p}\text{where}t\in \mathbb{R}$$

• $\vec{d}$ Is the directional vector for $\ell$, ”for some” indicates that the parameter variable $t$ would be different, but specific for each line.

  • $\forall$ This would make the statement false because x cannot be the same for all t in R

• Parameter variable is a dummy variable, does not need to solve it. But if if appears in 2 different vector forms, it doesn’t mean they are equal

  • Use different parameter variable when comparing different vector forms

• $\{\begin{array}{cc}& \vec{x}=t\vec{d}+\vec{p}\text{for some}t\in \mathbb{R}\\ & \vec{x}=t\vec{d}+\vec{p}\text{for all}t\in \mathbb{R}\end{array}$ $\begin{array}{c}\text{are either True or False}\\ \text{they did not specify}\ell \text{in vector form}\end{array}$

• Vector lines is a set, but a vector (it contains many vectors). Vector form is a shorthand for a set, but it is not equivalent to a set

  • Therefore for two vector lines to be equal, iff ${\ell }_1\subseteq {\ell }_2\text{and}{\ell }_2\subseteq {\ell }_1$ is True

  Show ${\ell }_1$ and ${\ell }_2$ are the same line (vector forms $\vec{x}=t\left[\begin{array}{c}1\\ 1\end{array}\right]+\left[\begin{array}{c}2\\ 1\end{array}\right]$ and $\vec{x}=t\left[\begin{array}{c}2\\ 2\end{array}\right]+\left[\begin{array}{c}4\\ 3\end{array}\right]$ ) §

  WTS: $\forall \vec{x}\in {\ell }_1\Rightarrow \vec{x}\in {\ell }_2$ (in another words, if ${\ell }_1={\ell }_2$ always has a solution)

  Set parameter variables as $t$ and $s$

  $$t\left[\begin{array}{c}1\\ 1\end{array}\right]+\left[\begin{array}{c}2\\ 1\end{array}\right]=\vec{x}=s\left[\begin{array}{c}2\\ 2\end{array}\right]+\left[\begin{array}{c}4\\ 3\end{array}\right]$$

  Solving this vector equation:

  $$\left[\begin{array}{c}-2\\ -2\end{array}\right]=s\left[\begin{array}{c}2\\ 2\end{array}\right]-t\left[\begin{array}{c}1\\ 1\end{array}\right]=\vec{0}=(s+1-\frac{t}{2})\left[\begin{array}{c}2\\ 2\end{array}\right]$$

  Since $s$ always has a solution for any $t$ AND $t$ always has a solution for any $s$, ${\ell }_1$ and ${\ell }_2$ is equal

Vector Form in Higher Dimensions §

• Skew - lines that does not intersect even if their directional vectors are different
• Overdetermined system - there are more equations than the number of unknown variables
• Underdetermined system - there are less equations than the number of unknown variables

Vector Plane §

• Needs 3 distinct points which are not on the same line to define a plane

• For some vectors ${\vec{d}}_1$ and ${\vec{d}}_2$ and point $\vec{p}$

  $$\mathcal{P}=\{\vec{x}\mid \vec{x}=t{\vec{d}}_1+s{\vec{d}}_2+\vec{p}\text{for some}t,s\in \mathbb{R}\}$$

• ${\vec{d}}_1$ And ${\vec{d}}_2$ are both the directional vectors for $\mathcal{P}$, extended out of the same point

• $\vec{x}=t{\vec{d}}_1+s{\vec{d}}_2+A=t(\overrightarrow{AB})+s(\overrightarrow{AC})+A=t(\vec{b}-\vec{a})+s(\vec{c}-\vec{a})+\vec{a}$

  Find the Line of Intersection between ${\mathcal{P}}_1$ and ${\mathcal{P}}_2$ with planes: $\vec{x}=t\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]+s\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]+\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$ And $\vec{x}=t\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]+s\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 3\end{array}\right]$ §

  Line of Intersection means that ${\mathcal{P}}_1={\mathcal{P}}_2$, and set the parameter variables as $t,s$ and $n,m$

  $$\vec{x}=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]=-t\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]-s\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]+n\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]+m\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]$$

  Equivalent to the system:

  $\{\begin{array}{ccc}& -n+m-t+s& =1\\ & 2m-t& =2\\ & 2n+m-s& =0\end{array}$ Thus: $\left[\begin{array}{c}t\\ s\\ n\\ m\end{array}\right]=\left[\begin{array}{c}2(r-2)\\ r-2\\ -1\\ r\end{array}\right]$

  Substitute the values back to vector form in ${\mathcal{P}}_2$

  $$\vec{x}=r\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]+\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$$

  This is the vector form for the line of intersection between ${\mathcal{P}}_1$ and ${\mathcal{P}}_2$

  Describe the plan $\mathcal{P}\subseteq {\mathbb{R}}^3$ with equation $z=2x+y+3$ in vector form §

  Get points on $\mathcal{P}$, $A=\left[\begin{array}{c}0\\ 0\\ 3\end{array}\right]B=\left[\begin{array}{c}1\\ 0\\ 5\end{array}\right]C=\left[\begin{array}{c}0\\ 1\\ 4\end{array}\right]$

  Thus, the directional vectors are,

  $${\vec{d}}_1=B-A=\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]\text{and}{\vec{d}}_2=C-A=\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]$$

  Since these vectors are not parallel, $\mathcal{P}$ can be expressed in vector form as

  $$\vec{x}=t{\vec{d}}_1+s{\vec{d}}_2=t\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]+s\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 3\end{array}\right]$$

Restricted Linear Combinations §

• Sets can be restricted to display rays, segments, and different shapes
• Different shapes of vectors on the graph can be made with restricting the coefficients of their linear combinations

Non-negative linear combination §

• Let $\vec{w}={\alpha }_1{\vec{v}}_1+{\alpha }_2{\vec{v}}_2...+{\alpha }_n{\vec{v}}_n$, it would be a non-negative linear combination of ${\vec{v}}_1,{\vec{v}}_2...{\vec{v}}_n$ if ${\alpha }_1,{\alpha }_2...{\alpha }_n\ge 0$ (all parameters are bigger or equal to 0)
• Vectors that can only displacing “forward”

Convex linear combination §

• Let $\vec{w}={\alpha }_1{\vec{v}}_1+{\alpha }_2{\vec{v}}_2...+{\alpha }_n{\vec{v}}_n$ , it would be a convex linear combination of ${\vec{v}}_1,{\vec{v}}_2...{\vec{v}}_n$ if

  $${\alpha }_1,{\alpha }_2...{\alpha }_n\ge 0\wedge {\alpha }_1+{\alpha }_2+...+{\alpha }_n=1$$

• The weighted averages of

• vectors (the average of ${\vec{v}}_1,...{\vec{v}}_n$ would be convex linear combination with coefficients ${\alpha }_i=\frac{1}{n}$)s