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

[Fernando Revilla]

Here is the question:

Find parametric equations of the line of intersection of the two planes 7x+8y = -1 and -9x-7y+4z = -7.
What is the direction vector used?

Here is a link to the question:

Calculus III Intersecting Planes? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

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 $x=t$.