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Homework Help: 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


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    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


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    No, that is not the direction vector, it is the position vector.
  9. Oct 31, 2007 #8
    Great, thanks a lot.
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