New Notes Blog: Find the line of intersection between the two planes `z-x-y=0` and `z-2x+y=0` .

Wednesday, 17 June 2015

Find the line of intersection between the two planes $\displaystyle{z}-{x}-{y}={0}$ and $\displaystyle{z}-{2}{x}+{y}={0}$ .

First, know that the vector that is in the direction of the line of intersection $\displaystyle{\left({r}\right)}$ is the cross product between the normal vectors of the planes since it is perpendicular to each of them. A plane in the form $\displaystyle{a}{x}+{b}{y}+{c}{z}={0}$ has a normal vector $\displaystyle{n}=\lt{a},{b},{c}\gt$ .

Let $\displaystyle{z}-{x}-{y}={0}$ be plane $\displaystyle{1}$ and $\displaystyle{z}-{2}{x}+{y}={0}$ be plane $\displaystyle{2}$ . Then the two normal vectors are $\displaystyle{n}_{{1}}=\lt-{1},-{1},{1}\gt$ and $\displaystyle{n}_{{2}}=\lt-{2},{1},{1}\gt$ .

$\displaystyle{r}={n}_{{1}}\times{n}_{{2}}={\left(-{1}\cdot-{1}\right)}{i}-{\left(-{1}+{2}\right)}{j}+{\left(-{1}-{2}\right)}{k}=-{2}{i}-{j}-{3}{k}=\lt-{2},-{1},-{3}\gt$

A line has the...

$\displaystyle{r}={n}_{{1}}\times{n}_{{2}}={\left(-{1}\cdot-{1}\right)}{i}-{\left(-{1}+{2}\right)}{j}+{\left(-{1}-{2}\right)}{k}=-{2}{i}-{j}-{3}{k}=\lt-{2},-{1},-{3}\gt$

A line has the parametric equation

$\displaystyle{L}{\left({t}\right)}={\left({a},{b},{c}\right)}-{t}\cdot{r}$

Where point $\displaystyle{\left({a},{b},{c}\right)}$ is any point on the line in the direction of $\displaystyle{r}$ . So all we need now is a point on the line. The vector $\displaystyle{r}$ has a $\displaystyle{z}$ component of $\displaystyle-{3}$ which means at some value of t it must go through the plane $\displaystyle{z}={0}$ . Therefore, we can set $\displaystyle{z}={0}$ in the plane $\displaystyle{1}$ and $\displaystyle{2}$ equations then solve for the $\displaystyle{x}$ and $\displaystyle{y}$ coordinates.

$\displaystyle{\left({1}\right)}:\to{0}-{x}-{y}={0}$

$\displaystyle{\left({2}\right)}:\to{0}-{2}{x}+{y}={0}$

Solving these equations gives $\displaystyle{x}={0}$ and $\displaystyle{y}={0}$ . So a point that the plane goes through is $\displaystyle{\left({0},{0},{0}\right)}$ . Then an equation has for the line must be