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    Two planes have vector equations r.(2i+j-3k)=3 and r.(2i+j-3k)=9

    Find the distance between them. I'm not too sure how to do this, as I haven't been given a point on the planes?
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    Well, note that the plane's normals are the same XD
    EDIT: Oh, derp, of course the normals are the same. I obviously need to start reviewing vectors!

    Also, if not having a point were the problem, you could just find one. For the first one, there's (0,0,-1), (1,1,0), etc...
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    (Original post by Music99)
    Two planes have vector equations r.(2i+j-3k)=3 and r.(2i+j-3k)=9

    Find the distance between them. I'm not too sure how to do this, as I haven't been given a point on the planes?
    ^As aznkid66 said. Or, you also have the distance of the planes from the origin and you could find the difference between this as the planes are parallel.
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    (Original post by Felix Felicis)
    ^As aznkid66 said. Or, you also have the distance of the planes from the origin and you could find the difference between this as the planes are parallel.
    Makes sense. So with the distance of the plane from the origin its 9 and 3, so the difference is 6, but then what?
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    (Original post by Music99)
    Makes sense. So with the distance of the plane from the origin its 9 and 3, so the difference is 6, but then what?
    The distance from the origin isn't 9 and 3. If you have a plane in scalar product form \mathbf{r} \cdot \mathbf{n} = D then the distance from the origin is \dfrac{D}{| \mathbf{n} |}
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    (Original post by Felix Felicis)
    The distance from the origin isn't 9 and 3. If you have a plane in scalar product form \mathbf{r} \cdot \mathbf{n} = D then the distance from the origin is \dfrac{D}{| \mathbf{n} |}
    Ah yeah, I keep getting confused and thinking i'm dealing with the unit normal vector :P.

    Also how would you convert this into scalar product form?

    x=-1 +2K + 3D
    y=2-K-3D
    z=-2+4K+5D

    I tried to eliminate K and D but I must be doing it wrong...
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    Um...what?

    it's already in scalar product form (x,y,z).(2,1,-3)=3 (or 9).

    Cartesian form would be multiplying everything out into 2x+y-3z=3 (or 9).

    But this is irrelevant to the question because the lines showing the distance from the origin to the plane are collinear. It's a one-dimensional problem - simply one distance minus the other.


    EDIT: Apparently I have no idea what I'm doing.
    EDIT2: Ah, I see, it was a different problem. My bad, my bad. ^^
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    (Original post by Music99)
    Ah yeah, I keep getting confused and thinking i'm dealing with the unit normal vector :P.

    Also how would you convert this into scalar product form?

    x=-1 +2K + 3D
    y=2-K-3D
    z=-2+4K+5D

    I tried to eliminate K and D but I must be doing it wrong...
    You have two lines in the plane:

    \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix} + K \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + D \begin{pmatrix} 3 \\ -3 \\ 5 \end{pmatrix}

    Use the cross product to find the normal vector and use the dot product of the position vector of one point on plane to find D
 
 
 
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