Planes in space/Equations 4x-2y-3z is 5, 3x-5y+2z is 1/Intersection line/Example

Two planes in space, intersecting in a line.

Suppose that two planes are given in ${}\mathbb {R} ^{3}$,

${}E={\left\{(x,y,z)\in \mathbb {R} ^{3}\mid 4x-2y-3z=5\right\}}\,$

and

${}F={\left\{(x,y,z)\in \mathbb {R} ^{3}\mid 3x-5y+2z=1\right\}}\,.$

How can we describe the intersecting line ${}G=E\cap F$? A point ${}P=(x,y,z)$ belongs to the intersection line if and only if it satisfies both plane equations. Therefore, both equations,

$4x-2y-3z=5\,\,{\text{ and }}\,\,3x-5y+2z=1,$

must hold. We multiply the first equation by ${}3$, and subtract from that four times the second equation, and get

${}14y-17z=11\,.$

If we set ${}y=0$, then ${}z=-{\frac {11}{17}}$ and ${}x={\frac {13}{17}}$ must hold. This means that the point ${}P=\left({\frac {13}{17}},\,0,\,-{\frac {11}{17}}\right)$ belongs to ${}G$. In the same way, setting ${}z=0$, we find the point ${}Q=\left({\frac {23}{14}},\,{\frac {11}{14}},\,0\right)$. Therefore, the intersecting line is the line connecting these points, so

${}G={\left\{\left({\frac {13}{17}},\,0,\,-{\frac {11}{17}}\right)+t\left({\frac {209}{238}},\,{\frac {11}{14}},\,{\frac {11}{17}}\right)\mid t\in \mathbb {R} \right\}}\,.$