Intersecting Lines and Planes

[Sparky09]

Hello everyone,
Thanks in advance for any help anyone can provide. I'm working on a self-study course through the adult learning center and they don't do a very good job of explaining things. I have been able to work through how to find the intersecting point of a line through a plane up until now. The problem now is that they have given me a problem that isn't explained. I've tried to work backwards to find the answer but cannot figure out what way to attack it. Here's the problem as given:

Determine the values of k such that the line x+2/-1=y-3/2=z-1/k and the plane 2x-4y+z+11=0 intersect at:
a) a single point
b) an infinite number of points
c) no point

I know normally to find the intersecting point you would find the parametric equations of the line and then substitute them into the cartesian equation of the plane, solve for t, then substitute t into the parametric equations to find the values for x,y,z. I know the parametric equations for the line would be:

x=-2-t y=3+2t and z=1+ kt

However, substituting these values into the cartesian equation at this point will do me no good because there are two unknowns .. t and k
So I thought since x+2/-1 is equal to y-3/2 .. I could substitute the parametric equations into the original symmetric equation and find the value for t .. then substitue t into z-1/k =x+2/-1 and solve for k that way. I don't think this is the right way to go and I've tried it several times but it doesn't work. I'm not even sure you CAN substitute a parametric equation into itself like that.. i imaine not. I know in order for the line to have an infinite number of points it means the plane's cartesian equation must be true no matter what the value of t ends up being and I also know that in order for it to not intersect at all the value of t can be anything and it will not equal out. I'm just not sure how to get from the symmetric equations to finding the value of t with the given information.

Helllp!!

 

[galactus]

Subbing the parametric equations into the plane equation gives:

\(\displaystyle 2(-2-t)-4(3+2t)+(1+kt)+11=0\)

\(\displaystyle t=\frac{4}{k-10}\)

There is a single intersection point if \(\displaystyle k\neq 10\)

No intersection if k=10

There are infinite solutions if the line lies in the plane. Since the above t equation is not true for any k, then there is no k that gives infinite solutions.

 

[soroban]

Hello, Sparky09!

After seeing galactus' excellent solution, I tried a different approach.

\(\displaystyle \text{Determine the values of }k\text{ such that the line: }\: \frac{x+2}{\text{-}1} \:=\: \frac{y-3}{2} \:=\:\frac{z-1}{k}\)

\(\displaystyle \text{and the plane }\,2x-4y+z+11\:=\:0\,\text{ intersect at: }\)

. . \(\displaystyle \text{(a) one point}\)
. . \(\displaystyle \text{(b) an infinite number of points}\)
. . \(\displaystyle \text{(c) no point}\)

\(\displaystyle \text{The line }L\text{ has direction vector: }\:\vec v \:=\:\langle \text{-}1,2,k\rangle\)

\(\displaystyle \text{The plane }P\text{ has normal vector: }\:\vec n \:=\:\langle 2,\text{-}4,1\rangle\)


\(\displaystyle \text{If (c) line }L\text{ and plane }P\text{ do not interset, then: }\:L \parallel P \quad\Rightarrow\quad \vec v \perp \vec n.\)
. . \(\displaystyle \text{Hence: }\:\langle \text{-}1,2,k\rangle \cdot \langle 2,\text{-}4,1\rangle \:=\:0 \quad\Rightarrow\quad -2 - 8 + k \:=\:0 \quad\Rightarrow\quad k \:=\:10\)

\(\displaystyle \text{Therefore: }\:\begin{Bmatrix}\text{(a) one point:} & k \:\ne\;10 \\ \text{(c) \;\:no point:} & k \:=\:10\end{Bmatrix}\)

\(\displaystyle \text{There are }no\text{ values of }k\text{ for case (b).}\)

 

[Sparky09]

Wow,
Thank you SO much both of you. It seems quite simple now that you've put it that way. I can't believe I racked my brain for an hour and a half and didn't think that up ha ha. Can't thank you enough. Happy to have some people that are good at math and so quick to respond with the help. And I'm off to Unit 10! Have a great day everyone and thanks again
Sparky