Ok so this is basically for my upcoming test. The essence of question is that I have a plane $$E_1: \; 18x-y-10z=1$$ and a line $$g(x)=(1,-3,-2)+x(1,-2,0)$$
What I need to do next is find a plane $E_2$ which is perpendicular to $E_1$ and containing the line described by $g(x)$
Since this is a part of a bigger problem the plane $E_1$ already contains the line $g(x)$
I have tried to do following. Since the line is given I can find any two points which it contains. So I choose $x=0$ and $x=1$ and plug them into function to obtain two points contained by the line. These points are:
$$P_0(1,-3,-2)$$
$$P_1(2,-5,-2)$$
Now since I have two points,I can find a vector between them $$P_0P_1 (1,-2,0)$$
Also I can determine the normal vector of plane $E_1$ which will be denoted $n_1$ and is equal to $$n_1=(18,-1,-10)$$
Now let $n_2$ be the perpendicular vector of plane $E_2$.Since these two planes are perpendicular to each other,then their normal vectors must be perpendicular to each other, and dot product of the normal vectors of two planes must be zero. When I take scalar product I get the first equation on my way:
$$18x-y-10z=0$$
Next equation I obtain is by taking in consideration that $n_2$ is normal to plane $E_2$ thus normal to contained vector $P_0P_1$ which gives the second equation:
$$x-2y=0$$
From here on I am so to say, stuck. I cant obtain a second equation, nor can I find a third non-collinear point to finish the system. How do I determine the plane?
Best Answer
You have two vectors parallel to $E_2$: $v=(1,-2,0)$ and $n_1=(18,-1,-10)$, so you can write immediately the parametric equations for $E_2$: $$ (x,y,z)=P_0+sv+tn_1=(1+s+18t,-3-2s-t,2-10t). $$ The cartesian equation can be readily found from these: express $t$ and $s$ using the equations for $z$ and $y$, then substitute into the equation for $x$.