When given a line in a plane and a line parallel to a plane I am expected to use these to find the normal to the plane as follows:
If plane is $ax + by + cz = d$ where $(ai + bj + ck)$ is the normal to the plane and direction vectors of two aforementioned lines are $(pi + qj + rk)$ and $(si + tj + uk)$.
Then $pa + qb + rc = sa + tj + uc = 0$ and from this you can get a numerical ratio for $a:b:c$.
An example would be $a + 2b + 3c = 0$ and $2a + b -2c = 0$
$a = -2b - 3c$ so $2(-2b -3c) + b - 2c = 0$
$b = -(8/3) c$ so $a = (7/3) c$
leading to ratio $a:b:c = 7:(-8):3$
You are then told to use this ration as the values of $a, b$ and $c$ i.e. $7x - 8y + 3z = d$
Using a point in the plane it is then easy to find $d$.
My question is why can you use the ratio as the values for the normal? In doing so one makes a presumption which could be wrong.
In the case of my example the presumption is that $c = 1$.
Hint:
all the vectors $(7k,-8k,3k)^T$ for $k \ne 0$ are parallel and can be used as orthogonal vectors to the plane.