- #1

singleton

- 117

- 0

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!

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!

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