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John's question at Yahoo! Answers (parametric equations of a line).

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello John,

A direction vector of the intersection line $r$ is

$v=(7,8,0)\times (-9,-7,4)=\det \;\begin{bmatrix}{i}&{j}&{k}\\{7}&{8}&{0}\\{-9}&{-7}&{4}\end{bmatrix}=(32,-28,23)$

For $x=0$ we get the system $8y=-1,\;-7y+4z=-7$ which implies $y=-1/8$ and $z=-63/32$. Hence,

$r\equiv\left \{ \begin{matrix}x=32t\\y=-\dfrac{1}{8}-28t\\z=-\dfrac{63}{32}+23t\end{matrix}\right.\quad (t\in\mathbb{R})$


Another way is to solve the system:

$\left \{ \begin{matrix}7x+8y = -1\\-9x-7y+4z = -7\end{matrix}\right.$

as a function (for example) of $x=t$.