Hi
Can someone tell me how to do the following question:
Find the parametric equation of the straight line of intersection of the planes:
$\displaystyle -2x+11y+4z=-1$
$\displaystyle -x+2y+z=3$
P.S
This question is not advanced algebra but analytic geometry.
1) Solve the system of two equations determined by the planes' eq's to find the intersection of the two planes.
2) Now parametrize the line you found in (1).
Do the above and if you get stuck somewhere write back.
Tonio
Input the above in the first eq. and you'll see it is completely wrong...
Multiply the 2nd. equation by -2 and add to this the 1st. one to obtain $\displaystyle 7y+2z=-7\Longrightarrow y=-1-\frac{2}{7}z$ , and inputting
this in the 2nd. original eq. we get $\displaystyle -x-2-\frac{4}{7}z+z=3\Longrightarrow x=-5+\frac{3}{7}z$ , and thus the intersection line is given
by $\displaystyle \left\{\begin{pmatrix}-5+\frac{3}{7}t\\{}\\-1-\frac{2}{7}t\\{}\\t\end{pmatrix}\,,\,t\in\mathbb{R }\right\}$ .
The above is already a parametrization, but your teacher may want you to write it as $\displaystyle u+tv\,,\,u,v$ vectors
on the line, $\displaystyle t=$ the real parameter.
Tonio