Calculus: Lines and Planes in Space - question

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caffeine206

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The line of intersection of two perpendicular planes is given by the parametrizationL(t) = (1 + t; 1 + 2t; 1 + 3t)
One of the planes goes through the origin. Find the equations of both planes.

I am thrown off by this question because the information they provide you with doesn't match other problems I've encountered. I'm guessing that the part about one of the planes going through the origin is important - we know that one of the planes travels through the point (0,0,0). The given line also tells us the point of intersection, (1,1,1), but I'm not sure exactly what relationship these two points have with each other.
 
For any t, the point P(t) = (1+t,1+2t,1+3t) lies in both planes. Let Plane 1 be the one containing the origin. Pick a t, say 0. Then (0,0,0), P(0) lie in plane 1. Use these points to get a vector, and take the cross product of the line's direction vector with it. This gives the normal N_1 to Plane 1, which allows you to get the equation for Plane 1.


For the second plane, since N_1 is perpendicular to Plane 1, and Plane 1 and Plane 2 are perpendicular, N_1 lies in Plane 2. Cross it with the direction vector of the line to get a normal for Plane 2.
 
The given line also tells us the point of intersection, (1,1,1)
Click to expand...
What do you mean by "the point of intersection"? You have two planes, intersecting in a line you are given. (1, 1, 1) is a point on that line so is a point of intersection- but so is any other point on that line.

What I would do is this: taking t= 0 gives (1, 1, 1) as a point on the line of intersection. Taking t= 1 gives (2, 3, 4) as another point of on the line of intersection. Since they are on the line of intersection, the vector <2- 1, 3- 1, 4- 1>= <1, 2, 3> as a vector in both planes. Since (0, 0, 0) is in one of the planes, <1- 0, 1- 0, 1- 0> is a vector in that plane. The cross-product of <1, 2, 3> and <1, 1, 1> gives a normal vector to the plane. Once you have that normal vector, and the point (0, 0, 0), you can find the equation of the plane.
 
Thanks for the quick response.

So if I use point A(0, 0, 0) and point B(1, 1, 1) to get vector AB, I would get (1-0), (1-0), (1-0) = i + j + k.

The cross product would then be: (1i + 1j + 1k) x (i + 2j + 3k) = (3-2) - (3-1) + (2-1) = i - 2j + k

By plugging in the coordinates for point A, I would get: x - 2y + z = 0 as my equation for plane 1.

Do I have this right so far?

EDIT: I am trying it with t=0 and t=1

EDIT2:

(1+t, 1+2t, 1+3t)
@t = 0, A(1,1,1)
@t = 1, B(2,3,4)
C(0,0,0)

AB = (1,2,3)
CA = (1,1,1)

AB x CA = -1i + 2j - 1k

Since the components are (0,0,0), does that mean that the equation for plane 1 is: -x + 2y -z = 0 ?
 
Last edited:
(1+t, 1+2t, 1+3t)
@t = 0, A(1,1,1)
@t = 1, B(2,3,4)
C(0,0,0)

AB = (1,2,3)
CA = (1,1,1)

AB x CA = -1i + 2j - 1k

Since the components are (0,0,0), does that mean that the equation for plane 1 is: -x + 2y -z = 0 ?
 
For some reason I was having trouble replying on my other account. How am I doing so far on this?
(1+t, 1+2t, 1+3t) @t=0 A(1,1,1) @t=1 B(2,3,4) C(0,0,0) AB = (1,2,3) CA=(1,1,1)
AB x CA = -1i + 2j -1k Does this make the equation for Plane 1: -x +2y - z (since the components for C are 0,0,0)?
 
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