1

Determine a vector equation of the line of intersection of the planes p1: $3x - y + 4z - 2 = 0$ and p2: $x + 6y + 10z + 8 = 0$.

I know that I can find the cross product of the normals of these vectors (direction vector), but for a line, i need a direction vector AND a point, how would I get the point? Or is there another way to do it?

Thanks in advance!

Edit:

I got:

[2 3 -1 | 4] ~ [2 3 -1| -4]

[1 0 3 | 5] [0 0 -7 | -14] R1 - 2R2

therefore, t = 2

then how do I find x and y?

mich95
  • 8,713
Ol' Reliable
  • 349

3 Answers3

1

Although there are many conventional methods to solve this problem I thought you might like to see how it is done with Grassmann algebra. Specifically this uses the code of John Browne incorporated in a developmental Mathematica Application.

We initialize the application and set Grassmann 3-space with coordinates x, y and z.

<< GrassmannCalculus`
SetPreferences["Grassmann3Space", "Vector"]

Define each plane by the three points of axis intersection. The Grassmann expression is the exterior product of the three points. [FivePointedStar][DoubleStruckCapitalO] is the origin.

enter image description here

enter image description here

The intersection is given by the regressive product of the two planes, simplified by CongruenceSimplify:

enter image description here

enter image description here

We can factor this into an exterior product of a point and a direction vector through the point. This defines the line of intersection. The overall numerical coefficient can be ignored.

enter image description here

enter image description here

We can check the answer by checking that the exterior product of the intersection with each of the planes is zero. This means that the intersection is indeed in each of the planes.

enter image description here

{True, True}
David Park
  • 183
  • What you did is wayyyyyyyy beyond where I am (Grade 12). But thanks for your help! – Ol' Reliable Oct 08 '13 at 04:28
  • The exterior product just keeps extending the dimension of an object, from a point to a line to a plane. A plane is defined by three points and I hope you see how the axis intersections were obtained. The regressive product is more abstract but it does give intersections of geometric "planes". And I hope you see how it does give you your answer directly in the form of a point and a vector - just as you asked. – David Park Oct 08 '13 at 04:39
1

You just need to choose a point that satisfies both $3x - y + 4z - 2 = 0$, and $x + 6y + 10z + 8 = 0$, knowing that any such point will be on the line of intersection. Choosing $x=2$ and solving the resulting system of equations \begin{align} -y+4z+4=0 \\ 6y+10z+10=0 \end{align} for $y$ and $z$, we see that $y=0$ and $z=-1$, so $(2,0,-1)$ is a point on the line of intersection.

stochasm
  • 1,607
  • would (-16,-13,9) also work? – Ol' Reliable Oct 08 '13 at 04:27
  • No, notice that substituting those values of $x$, $y$, and $z$ into $3x-y+4z-2=0$, we get $-1=0$. Remember that the point you choose must satisfy the equations of both of your planes. – stochasm Oct 08 '13 at 04:34
  • Ok, can you also check if this is correct: http://math.stackexchange.com/questions/518530/find-the-equation-of-the-plane-that-passes-through-the-line-of-intersection-of-t – Ol' Reliable Oct 08 '13 at 04:43
0

You could solve the system of equations \begin{align*} 3x-y+4z-2 &= 0\\ x+6y+10z+9 &= 0 \end{align*} by eliminating variables. Then you would have a parameterization for the line.

J126
  • 17,451