~~REVEAL~~ ==== Linear equations in 3 variables ==== ==== Definition ==== {{page>linear equation in 3 variables}} ==== Examples ==== * $x+y+z=1$ ==== ==== * $x+y=1$ This may be viewed as a linear equation in 3 variables, since it is equivalent to $x+y+0z=1$. ==== ==== * $z=1$, viewed as the equation $0x+0y+z=1$
This plane is horizontal (parallel to the $x$-$y$ plane). ==== Linear equations (in general) ==== {{page>linear equation}} ==== Example ==== \[ 3x_1+5x_2-7x_3+11x_4=12\] is a linear equation in 4 variables. * A typical solution will be a point $(x_1,x_2,x_3,x_4)\in \mathbb{R}^4$ so that $3x_1+5x_2-7x_3+11x_4$ really does equal $12$. * For example, $(-2,0,-1,1)$ is a solution. * The set of all solutions is a 3-dimensional object in $\mathbb{R}^4$, called a [[wp>hyperplane]]. * Since we can't draw pictures in 4-dimensional space $\mathbb{R^4}$ we can't draw this set of solutions! ==== Systems of linear equations ==== {{page>system of linear equations}} ==== Example ==== Find the line of intersection of the two planes $ x+3y+z=5$ and $ 2x+7y+4z=17$. * ==== Intersection of $ x+3y+z=5$ and $ 2x+7y+4z=17$ ==== * To find the equation of the line of intersection, we must find the points which are solutions of //both// equations at the same time. * Eliminating variables, we get $x=-16+5z$, $y=7-2z$ * The line of intersection consists of the points $(-16+5z,7-2z,z)$, where $z\in\mathbb{R}$ ==== A detailed look at the last example ==== * $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$ * Find solutions of this [[system of linear equations|system]] by applying operations * Aim to end up with a very simple sort of system where we can see the solutions easily. ==== ==== * $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$ * Replace equation (2) with $(2)-2\times (1)$: * $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$ * Now replace equation (1) with $(1)-3\times (2)$ * $\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$ ==== ==== * $\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$ * can easily rearrange (1) to find $x$ in terms of $z$ * can easily rearrange (2) to find $y$ in terms of $z$ * Since $z$ can take any value, write $z=t$ where $t$ is a "free parameter" * (which means $t$ can be any real number, or $t\in \mathbb{R}$). * Solution: \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*} ==== ==== * Solution: \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*} * Can also write this in "vector form": * $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}.$ * This is the equation of the line where the two planes described by the original equations intersect. ==== ==== * $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}$ * For each value of $t$, we get a different solution (a different point on the line of intersection). * e.g. take $t=0$ to see that $(-16,7,0)$ is a solution * take $t=1.5$ to see that $(-16+1.5\times 5,7+1.5\times (-2),1.5) = (-8.5,4,1.5)$ is another solution * etc. * This works for any value $t\in\mathbb{R}$, and every solution may be written in this way.