Normal between two lines

Question

$$l_1:x=(-1,1,2)+t(1,-3,-2)$$ $$l_2:y=(1,1,1)+s(-2,6,-4)$$ Q: Find the equation of plane passing through $l_1$ and $l_2$.

My teacher wrote answer like this:

Since they're parallel, choose vector $u=(1,-3,-2)$ and choose normal vector $n=u$x$(x_0-y_0)=(-3,-5,6)$. Can someone please explain me how did he find $n=(-3,-5,6)$.

No matter how I choose vectors $x_0,y_0$ it gives $n=(-3,3,-6)$. Is there a way to obtain $n=(-3,-5,6)$?

Thanks!

Answer 1

My teacher assigned us this problem in my multivariable calculus class and a classmate and I came to the conclusion that there must be a typo in the problem, because the two lines are skew.
Using the two lines $$ _1:=(−1,1,2)+(1,−3,−2) $$ $$ _2:=(1,1,1)+(−2,6,−4) $$ We can prove these are skew by first noticing the slope vectors are linearly independent, so the lines cannot be parallel. Next, to prove the lines don't intersect either, we parameterize the two lines to $l_1(t)=(t-1,-3t+1,-2t+2)$ and $l_2(t)=(-2t+1,6t+1,-4t+1)$. Then we can set each $x(t)$, $y(t)$, and $z(t)$ to the other to find where they intersect. This gives 3 $t$ values, $t=2/3,0,-1/2$. Because the 3 $t$ values are not equal, there isn't any one point where the two lines intersect, so the lines are skew.

Alternatively, if we use lines that we assume are the intended lines, $$ _1:=(−1,1,2)+(1,−3,−2) $$ $$ _2:=(1,1,1)+(−2,6,4) $$ These are parallel and we can find a plane that passes through both lines. First we have to find 4 points and subtract the difference to produce 2 vectors later used to find the normal of the plane. The two vectors in between the two lines, when crossed, will produce the vector normal.
To find 4 points, we can first set $t$ and $s$ to 0, and then to 1. We get $(-1,1,2)$ and $(1,1,1)$ when $t,s=0$ and $(0,-2,0)$ and $(-1,7,5)$ when $t,s=1$.
Subtracting, we get two vectors $(-2,0,1)$ and $(1,-9,-5)$. Now to find the normal, we cross the two vectors and get $n=(9,-9,18)$, which is parallel to $(−3,3,−6)$.
Now we use $n=(A,B,C)$ and one of the initial points (we'll use $(1,1,1)$ for simplicity) and plug them into the equation for the plane passing through both lines, $Ax + By + Cz = n \cdot $(1,1,1)$ $
This leaves us with $9x - 9y +18z = 18$.

Answer 2

The vector joining points A(-1,1,2) and B(1,1,1) is $\vec L=-2i+k$ which lies on the plane. Another vector lying in the plane is given as $\vec U=i-3j-2k$ so the normal to the plane is given by $\vec N=\vec L \times \vec U=3i-3j+6k$. So the equation of plane through the point B can be written as $\vec N.[\vec r-\vec B]=0 \implies (3i-3j+6k).\vec r=\vec N.\vec B \implies 3x-3y+6z=3-3+6 \implies x-y+2z=2.$

OP's $\vec N$ is correct.

Answer 3

• First of all, we know that the two lines l1,l2 are not parallels.

  We know that because, by definition two lines are parallels if their direction vectors are parallels, that means that the two vectors must be linearly dependents (or the same).

  In this case, the direction vector of l1 is v1 = (1, -3, -2) (the element of the equation that multiplies "t") and the direction vector of l2 is v2 = (-2, 6, -4). If we study these two vectors we see that they are not parallels because if we multiply v1 by -2 (to obtain the first component of v2, -2) we have as result 2*v1 = (-2, 6, 4) that is not v2, so we have proven that v1 and v2 are not linearly dependents so we have proven that l1 and l2 are not parallels.

• To find the equation of the plane we need the normal vector of the plain.

  By definition, the normal vector to a plane is a perpendicular vector to this plane.

  If we choose l1 to find the normal vector, we need to find a perpendicular vector of

                                 v1 = (1, -3, -2)
  

  Any perpendicular vector we choose will be fine. Your teacher has chosen n = (-3, -5, 6) but you can choose any perpendicular vector.

  To understand how n is obtained, you can fix for example the first and second components of n = (-3, -5, x) and try to find the value of x knowing that it has to happend that

  nv1' = 0 => (-3, -5, x) * (1, -3, -2)' = 0 => -3·1 + (-5)·(-3) + x·(-2) = 0 => -3+15-2x = 0
  => 12 - 2x = 0 => 12 = 2x => x = 6

  And finally we have that n = (-3, -5, 6) and that's the way we obtain this vector

I hope it works to you, any doubt I will glad to answer if I know it ;)