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?
3 Answers
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.
HINT
- determine a vector $n=(a,b,c)\perp q$
- consider the plane $ax+by+cz+d=0 \quad \parallel \quad L1$
- consider a point $Q\not \in L1$
- find $d$ by the condition $Q\in$ plane
the line $L_1$
- Passes through the point $P=(1,−2,0)$.
- Is parallel to the vector $\vec q = (1,−1,3).$
Find a plane that will have no points of intersection with the line $L_1.$
Create a vector perpendicular to $\vec q$, say $\vec v =(1,1,0)$. (Note $v \circ q = 0.)$
Then the point $Q = v+P = (2,-1,0)$ is not on the line $L_1$. (Note $P \ne Q.$)
The equation of the plane, $\pi$, that is perpendicular to the vector $\vec v$ and passes through the point $Q$, is
\begin{align} [(x,y,z)-Q] \circ \vec v &= 0 \\ (x-1) + (y+2) &= 0 \\ x+y &= -1 \end{align}
Since the plane and the line are both perpendicular to the line $\overleftrightarrow{PQ}$ at points $P$ and $Q$ respectively, then the line and the plane are parallel.
