So I've been stuck on this question.
Find a plane that will have no points of intersection with the line $L1.$
$L1$ passes through the point $P(1,−2,0)$ and parallel to the vector $q = (1,−1,3).$ Find a plane that will have no points of intersection with the line $L1.$
So straight away I think, find a plane parallel to the line $L1$
But the normal to a plane $(Ax+By+Cz)$ is $(A,B,C)$ if I've got that right.
Where is the hole in my knowledge? And how would I go about this correctly?
Best Answer
Choos any vector not parallel to $(1,-1,3)$ such as $(1,0,0)$
The vector perpendicular to both of these, i.e. The cross product is $(0,3,1)$
This is the normal to the plane.
The equation of the plane is therefore $3y+z=d$
Now we choose $d$ so that the point $(1,-2,0)$ does not lie on the plane, such as $d=0$ and you are done.