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Math Help - Intersecting Planes

  1. #1
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    Intersecting Planes

    Hello. Could somebody please help me out on some questions about intersecting planes? This is rather urgent so if you could help me out quickly, it'll be great.

    -----

    Two vectors a and b are the normals to two planes. Both of these planes pass through (0,0,0). Vector OP = t[a x b] where t is a scalar.

    1. Describe the set of all points P in relation to the two distinct planes

    For this one, can somebody perhaps explain it in "simpler" terms because I'm rather confused on the wording of it.

    2. From above or otherwise, find the parametric equations of the line of intersection between the two planes

    5x + 2y - z = 0
    3x + y + 4z = 0


    Okay, first, I got the direction vectors of the two equations which are [5,2,-1] and [3,1,4].

    From there, I cross-producted these two vectors to get [7,-17,-1] which is the perpendicular of the two vectors. Now vector OP is [7t,-17t,-t] where t is a scalar. Therefore P is (7t,-17t,-t). Also, (7,-17,-1) lies on both planes and hence, lies on the line of intersection.

    So I have two points (0,0,0) and (7,-17,-1) which lie on the line of intersection.

    Is what I did above correct? If so, what would be the next step to take?

    3. Show the line of intersection from before is parallel to plane 4x + y + 11= 26

    I'll need to do the above question first to answer this one so once I get an answer for it, I'll attempt to do this one.

    -----

    Thankyou, all help is appreciated.
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  2. #2
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    Quote Originally Posted by sqleung View Post
    ...
    Two vectors a and b are the normals to two planes. Both of these planes pass through (0,0,0). Vector OP = t[a x b] where t is a scalar.

    1. Describe the set of all points P in relation to the two distinct planes

    For this one, can somebody perhaps explain it in "simpler" terms because I'm rather confused on the wording of it. : All points P form the intersection line between the two planes. See attachment.

    2. From above or otherwise, find the parametric equations of the line of intersection between the two planes

    5x + 2y - z = 0
    3x + y + 4z = 0


    Okay, first, I got the direction vectors of the two equations which are [5,2,-1] and [3,1,4]. ... OK

    From there, I cross-producted these two vectors to get [7,-17,-1] wrong [9, -23, -1]

    ...
    to #2: The intersection line passing through (0, 0, 0) is:

    \vec r = t \cdot [9, -23, -1]

    to #3: The points P which are located on the intersection line have the coordinates P(9t, -23t, -t). Show that all points P have the same distance to the given plane.
    Attached Thumbnails Attached Thumbnails Intersecting Planes-schnittgerade_3ebenen.gif  
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  3. #3
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    Guess I made little mistake in my cross product...

    1. Thanks for putting it into simpler terms but how will I go about "describing" it? Will I have to draw a diagram?

    1. So form what I've gathered, it's the direction vector [9,-23,-1] through the point (0,0,0). With this, I get the equation:

    x = 9t
    y = -23t
    z = -t

    Is that right?

    3. Sorry but I'm not too sure how I can go about this one. Could you perhaps show me?

    -----

    Sorry for being such a pain. I'm just not really understanding this.

    Thanks for your help however.
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  4. #4
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    Quote Originally Posted by sqleung View Post
    Guess I made little mistake in my cross product...

    1. Thanks for putting it into simpler terms but how will I go about "describing" it? Will I have to draw a diagram?

    1. So form what I've gathered, it's the direction vector [9,-23,-1] through the point (0,0,0). With this, I get the equation:

    x = 9t
    y = -23t
    z = -t

    Is that right? .... OK

    3. Sorry but I'm not too sure how I can go about this one. Could you perhaps show me?

    ...
    to #1: The sentence in red in my previous post is the description. You can only add that the line of intersection must be perpendicular to the normal vectors of the planes.

    to #3: A line is parallel to a plane if the dot-product of the normal vector of the plane and the direction vector of the line equals zero, that means they must be perpendicular.

    1. I assume that the equation of the plane is: 4x + y + 11z = 26

    2. If you calculate [4, 1, 11] \cdot [9, -23, -1] = 2

    which indicates that the statement of your question is false.

    But if the two planes are

    5x + 2y +z = 0 and 3x + y + 4z = 0

    then the line of intersection would be:

    \vec r = t \cdot [7, -17, -1]

    and now this line is indeed parallel to the given plane because

    [4, 1, 11] \cdot [7, -17, -1] = 0
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  5. #5
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    Thankyou very much for that.

    I guess I typed in the wrong plane equation in the first place and so, my cross-product appeared wrong.

    Anyway, thanks!
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