# Determine the value of k using a line and plane

1. Apr 27, 2007

### Styx

Determine the value 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
no point

x = -2 - t
y = 3 + 2t
z = 1 + k

2(-2-t) -4(3+2t) + (1+k) +11 = 0
-4 - 2t - 12 -8t + 1 +k +11 = 0
10t - k = -4

For part a) I have concluded that k = and real number, k can not equal -10 or ...

For part c) I have concluded 10t - 10t = -4, 0t = -4
The planes do not intersect if k = -10

What I am having trouble with is part b). I have not sure how I would reach 0t = 0 which would put the line on the plane and therefore create infinite points of intersection

2. Apr 27, 2007

### HallsofIvy

Staff Emeritus
Well, you mean z= 1+ kt. I presume that was a typo.

Well, apparently it wasn't! You should have 2(-2-t)- 4(3+ 2t)+ (1- kt)+ 11= 0 or (-2t- 8t- kt)+ (-4- 12+ 1+ 11)= (-10+k)t- 4= 0.

There exist a single point of intersection if (-10+k)t= 4 has a unique solution: that is if -10+ k is not 0. That is, if k is not 10. (The equation you give "0t= -4" has no solution no matter what k is.)

There will be an infinite number of solutions if and only if the line is in the plane- that is, if x= -3-t, y= 3+2t, and z= 1+kt satisfies the equation of the plane for all t. Since putting that x, y, z into the equation of the plane reduces to (-10+ k)t= 4 which is NOT true for all t for any value of k (an equation of the form (-10+ k)t= 0 would be true for all t for k= 10) there is no value of k such that there is an infinite number of solutions.

If k= 10, then the equation reduces to 0*t= 4 which is never true and so there is no solution (the line is parallel to the plane).

3. Apr 27, 2007

### Styx

Sorry, but I am still a bit fuzzy on parts a and b

I had by signs reversed so for part c, k = 10, not -10.

As far as I can see for part a) k can equal any real number except 10, and whatever the answer to part b) is.

Part b) appears to have no solution though...

Last edited: Apr 27, 2007
4. Apr 28, 2007

### HallsofIvy

Staff Emeritus
I believe that was what I said in my first response:
"there is no value of k such that there is an infinite number of solutions."

5. Apr 28, 2007

### Styx

Thanks. I just wanted to make sure I wasn't misunderstanding you.