Intersection of a line and a plane, for what value(s) of k ?

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  • #1
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Main Question or Discussion Point

Well,

From what I understand, to determine the intersection of a line and a plane, we use parametric form of the line and substitute the values of x, y and z into the Cartesian equation of the plane, correct?

so, given the line
x = 2 + 4t
y = -1 + kt <=== note the 'k' variable
z = 5 - 3t

and the plane 7x + 6y - z - 3 = 0

What must be the value of k for no intersection point, one intersection point, an infinite number of intersection points?


*** My calculations this far are written below. So far I think I'm headed completely in the wrong direction but I've exhausted the only way I thought possible.

If I plug the parametric form into the plane equation, I end up with
31t + 6kt = 0

this is where I'm lost. The question has me lost, because I have not tried this form (usually I have been given a non-variable value for the parametric equations, plugging them in is easy and I go from there)

At this point, I'm -guessing- that to have no intersection point, I must have a constant on the other side of the equation and be inconsistent (of the sort, 0t = 123). This is not possible as I can't just invent one? So, there cannot be NO intersection (it must pass through the plane)

To have one intersection point, we can suppose any value of k, then t = 0?

And for infinite number of intersection points, basically I need 0t = 0 (dependent system)

SO, by letting k = - 31/6
31t + 6kt = 0
0t = 0
and thus an infinite number of intersection points (the line is contained in the plane)


*** The above is probably incorrect ( I feel) but I do not know another way. Please suggest the correct way of going about this. Do not give me the answer, I would rather just some advice on how to work my way back on this question! :smile:
 
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Answers and Replies

  • #2
TD
Homework Helper
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Your line is
[tex]\left( {\begin{array}{*{20}c}
x \\
y \\
z \\

\end{array}} \right) = \left( {\begin{array}{*{20}c}
2 \\
{ - 1} \\
5 \\
\end{array}} \right) + t\left( {\begin{array}{*{20}c}
4 \\
k \\
{ - 3} \\
\end{array}} \right)[/tex]

Note that the point (2,-1,5) lies in the plane, which is why you can't find a value of k for which there is no intersection.

For infinite intersections, the line has to be in the plane and k = -31/6 is correct there.
 
  • #3
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aha!

sorry, I'm slow on the uptake :(

thanks for the help :D
 
  • #4
TD
Homework Helper
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You do understand now, I hope? If there's something unclear, don't hesitate to ask!

Glad that I could help :smile:
 
  • #5
117
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Yeah, unfortunately I didn't think to look at the initial vector of the line being on the plane. I should have noticed that when I expanded the substitution and it came out to zero with the t variable left over... LOL

thanks again
 

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