1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Intersection of two planes

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Two planes [itex]r_1[/itex] and [itex]r_2[/itex] have the equations:

    [itex]r_1 = ( 1 - \lambda ) \underline{i} + ( 2 \lambda + \mu ) \underline{j} + ( \mu - 1 ) \underline{k}[/itex]

    [itex]r_2 = ( s - t ) \underline{i} + ( 2s - 3 ) \underline{j} + ( t ) \underline{k}[/itex]

    If a point lies in both [itex]r_1[/itex] and [itex]r_2[/itex] then [itex]\mu =4 \lambda + 3[/itex] (shown in a previous question)

    Hence find a vector equation of the line of intersection of the two planes.

    2. Relevant equations

    None known

    3. The attempt at a solution

    I know what I have to do but I have no idea how to do it:
    • Find the normals of the planes
    • Use the cross (vector) product on them to get the direction of the intersection vector
    • find a point on the vector (I assume using the [itex]\mu = 4 \lambda + 3[/itex] stuff)
    • substitute the two parts into the formula for a vector equation to get the answer
    However, I have no idea how to find the normals of those planes and I can also see finding the point to be awkward too with all those mu's, lambda's, s's and t's.
    I'm just completely stumped!
  2. jcsd
  3. Oct 29, 2007 #2
    Put the planes in the form Ax +By +C=0, eliminating the parameters in so doing, and the normals will be (A,B,C)
  4. Oct 30, 2007 #3
    The only way I know of making that form is by doing the dot product of the plane and its normal which doesn't help as the normal is what I'm trying to find :confused:
  5. Oct 30, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    If you have determined that [itex]\mu= 4\lambda + 3[/itex], that's all you need!

    Replace [itex]\mu[/itex] by [itex]4\lambda + 3[/itex] it the equation for the first plane:
    [itex]\vec{r_1}= (1- \lambda)\vec{i})+ (2\lambda + (4\lambda +3))\vec{j}+ ((4\lambda+ 3)- 1)\vec{k}[/itex]
    [itex]\vec{r_1}= (1-\lambda)\vec{i}+ (6\lambda+ 3)\vec{j}+ k+ (4\lambda+ 2)\vec{k}[/itex]
    That is the vector equation of the line satisfying [itex]\mu= 4\lambda+ 3[/itex]j- which you say is true for any point on the line of intersection.
  6. Oct 30, 2007 #5
    Thanks a lot, that helped loads!
  7. Oct 30, 2007 #6
    Sorry, but is that just the direction vector of the line of intersection?
    If so, then do I need to make it into the form [itex]r = \underline{a} + \lambda \underline{c}[/itex] ?
  8. Oct 31, 2007 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    No, that is not the direction vector, it is the position vector.
  9. Oct 31, 2007 #8
    Great, thanks a lot.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?