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?

  • $\begingroup$ Yes, that is correct. Now, are you trying to find a plane normal to the given line, as you say in your title, or are you trying to find a plane parallel to the line as you say in the body? A plane normal to the vector (1, -1, 3), containing the point (1, -2, 0) is (x- 1)- (y+ 2)+ 3z= 0. But there are an infinite number of planes parallel to a given line containing a given point! Consider the line the line through the given point parallel to the give line. Take any plane containing this second line. One of those planes contains the given line but all the rest are parallel. $\endgroup$ – user247327 Apr 6 '18 at 14:00
  • $\begingroup$ In order for there to be no point of intersection, the plane must be parallel to the given line. There are infinitely many possibilities $\endgroup$ – David Quinn Apr 6 '18 at 14:06

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.

  • $\begingroup$ Ahhhh, so by creating a line that isn't parallel to $L1$ we can find the normal to both with the cross product, thus the equation of the plane!! This makes alot of sense. $\endgroup$ – Dan D'silva Apr 6 '18 at 14:43


  • 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
  • $\begingroup$ So, I let some vector $v_1=(a,b,c)$ and do $v_1\cdot q=0$ to get $a-b+3c$ I don't understand where to go from there $\endgroup$ – Dan D'silva Apr 6 '18 at 14:16
  • $\begingroup$ take for example c=1b=1 a=4, we have infinitely many possibilities to choose that plane $\endgroup$ – gimusi Apr 6 '18 at 14:19
  • $\begingroup$ It seems to be that I'm confused as to how the answer should be. As I keep putting $ax+bz+cz=D$ And this always gives a plane perpendicular. Sorry if it sound dumb, but I really can't seem to get my head around it. $\endgroup$ – Dan D'silva Apr 6 '18 at 14:25
  • $\begingroup$ Choosing whatever $n=(a,b,c) \perp L1$ we obtain a family of planes parallel to L1 (as d varies), the it suffices to chose d in such way te plane doesn't contain L1 itself $\endgroup$ – gimusi Apr 6 '18 at 14:27

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.


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