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Math Help - Planes in 3-space; Equations of planes

  1. #1
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    Planes in 3-space; Equations of planes

    Hi guys,

    I've got this exercise where I'm required to describe the equation of the plane determined by the fact that the following
    three points lie in it: (−7, 1, 0), (2, −1, 3), (4, 1, 6).

    What i've done for it is as followings:

    To describe the equation of the plane we require a point in the plane and a vector n that is orthogonal to every vector in the plane. The vectorn is said to be normal to the plane.

    We suppose our point in the plane to be P(x0,y0, z0)
    We chose n the normal vector to be on of our given three points.
    n = (4, 1, 6)
    Now we have discovered a point within the plane and a normal we can carry out the equation of the form

    a(x-x0)+b(y-y0)+c(z-z0)
    where a,b,c = 4, 1, 6

    Well thats my attempt at the question i not quiet sure that i've answered it fully but i think i did most part correctly. I would appreciate any advice/help with it with the question it self and my answer (if it could be built on).

    Thanks a lot !
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  2. #2
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    You have made some mistake in calculations.
    Did you use the cross product?
    I get a=12,\;b=21,\;\&\,c=-22.
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  3. #3
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    Question

    Quote Originally Posted by Plato View Post
    You have made some mistake in calculations.
    Did you use the cross product?
    I get a=12,\;b=21,\;\&\,c=-22.
    Mistakes in which area or as whole ?
    erm...no i've not used cross product i wasn't aware i needed it to answer the question. So i should be using it and how?

    Thanks for the reply
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  4. #4
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    Quote Originally Posted by mathsToday View Post
    Mistakes in which area or as whole ?
    erm...no i've not used cross product i wasn't aware i needed it to answer the question. So i should be using it and how?

    Thanks for the reply

    You need to find two vectors that lie in the plane

    v_1= \underbrace{(-7,1,0)}_{Head}-\underbrace{(4,1,6)}_{tail}=-11 \vec i -6 \vec k

    and

    v_2= \underbrace{(2,-1,3)}_{Head}-\underbrace{(4,1,6)}_{tail}=-2 \vec i -2 \vec j -3 \vec k

    To find a normal vector we need to take the cross product

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
-11 && 0 && -6 \\<br />
-2 && -2 && -3 \\<br />
\end{vmatrix} = 12 \vec i +21 \vec j- 22 \vec k<br />

    This is the normal vector to the plane

    Good luck
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    You need to find two vectors that lie in the plane

    v_1= \underbrace{(-7,1,0)}_{Head}-\underbrace{(4,1,6)}_{tail}=-11 \vec i -6 \vec k

    and

    v_2= \underbrace{(2,-1,3)}_{Head}-\underbrace{(4,1,6)}_{tail}=-2 \vec i -2 \vec j -3 \vec k

    To find a normal vector we need to take the cross product

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
-11 && 0 && -6 \\<br />
-2 && -2 && -3 \\<br />
\end{vmatrix} = 12 \vec i +21 \vec j- 22 \vec k<br />

    This is the normal vector to the plane

    Good luck
    thanks a lot

    Oh ok it wasn't explained to me this way. And then so i use the found normal in the equation i proposed early ?
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  6. #6
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    ahh i get it a lot better now actually..think i may have answered it

    Thanks for help guys
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    You need to find two vectors that lie in the plane

    v_1= \underbrace{(-7,1,0)}_{Head}-\underbrace{(4,1,6)}_{tail}=-11 \vec i -6 \vec k

    and

    v_2= \underbrace{(2,-1,3)}_{Head}-\underbrace{(4,1,6)}_{tail}=-2 \vec i -2 \vec j -3 \vec k

    To find a normal vector we need to take the cross product

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
-11 && 0 && -6 \\<br />
-2 && -2 && -3 \\<br />
\end{vmatrix} = 12 \vec i +21 \vec j- 22 \vec k<br />

    This is the normal vector to the plane

    Good luck
    We the cross product i get 12i + 21k + 22k am i correct or have I carried out the wrong arithmetic?.
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    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
-11 && 0 && -6 \\<br />
-2 && -2 && -3 \\<br />
\end{vmatrix}<br />

    Expanding along the first row we get...

    i\cdot \begin{vmatrix}<br />
0 && -6 \\<br />
-2 && -3 \\<br />
\end{vmatrix} -j \cdot \begin{vmatrix}<br />
-11 && -6 \\<br />
-2 && -3 \\<br />
\end{vmatrix}+ k \cdot  \begin{vmatrix}<br />
-11 && 0 \\<br />
-2 && -2 \\<br />
\end{vmatrix}=

    (0(-3)-(-2)(-6)) \vec i -((-11)(-3)-(-2)(-6)) \vec j +((-11)(-2)-(-2)(0)) \vec k

    -12 \vec i -21 \vec j + 22 \vec k \iff 12 \vec i + 21 \vec j -22 \vec k

    Both of these are normal to the plane.
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  9. #9
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    Quote Originally Posted by TheEmptySet View Post
    You need to find two vectors that lie in the plane

    v_1= \underbrace{(-7,1,0)}_{Head}-\underbrace{(4,1,6)}_{tail}=-11 \vec i -6 \vec k

    and

    v_2= \underbrace{(2,-1,3)}_{Head}-\underbrace{(4,1,6)}_{tail}=-2 \vec i -2 \vec j -3 \vec k

    To find a normal vector we need to take the cross product

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
-11 && 0 && -6 \\<br />
-2 && -2 && -3 \\<br />
\end{vmatrix} = 12 \vec i +21 \vec j- 22 \vec k<br />

    This is the normal vector to the plane

    Good luck
    Furthering the equation i get

     12(x-x0) + 21(y-y0) -22(z-z0) = 0

    where as shown i obtained 12, 21 and -22 by calculating the normal. I then substitute a vector in the plane, i.e 4,1,6), giving you an equation

     12(x-4) + 21(y-1) -22(z-6) = 0.

    The equation can now be rewritten to the form
     ax+by+cz = d

    then
     12x+21y+-22z = d

    I'm not sure how to reorganize that equation to evaluate the d mentioned above. Any help /push in the right direction would be great!

    Once this is done then I should be able to check my equation by seeing whether each of the given three points in the plane satisfies it right ?

    Thanks a bunch
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  10. #10
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    Quote Originally Posted by mathsToday View Post
    Furthering the equation i get

     12(x-x0) + 21(y-y0) -22(z-z0) = 0

    where as shown i obtained 12, 21 and -22 by calculating the normal. I then substitute a vector in the plane, i.e 4,1,6), giving you an equation

     12(x-4) + 21(y-1) -22(z-6) = 0.

    The equation can now be rewritten to the form
     ax+by+cz = d

    then
     12x+21y+-22z = d

    I'm not sure how to reorganize that equation to evaluate the d mentioned above. Any help /push in the right direction would be great!

    Once this is done then I should be able to check my equation by seeing whether each of the given three points in the plane satisfies it right ?

    Thanks a bunch
     12(x-4) + 21(y-1) -22(z-6) = 0

    Distribute out the left hand side.

     12x-48 + 21y-21 -22z+132 = 0  \iff 12x +21y-22z +63=0

    so

    12x+21y-22z=-63

    P.S. if you want a sub script (underscore) use the following code

    <br />
12(x-x_0) + 21(y-y_0) -22(z-z_0) = 0

    12(x-x_0) + 21(y-y_0) -22(z-z_0) = 0
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  11. #11
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    Smile

    Quote Originally Posted by TheEmptySet View Post
     12(x-4) + 21(y-1) -22(z-6) = 0

    Distribute out the left hand side.

     12x-48 + 21y-21 -22z+132 = 0  \iff 12x +21y-22z +63=0

    so

    12x+21y-22z=-63

    P.S. if you want a sub script (underscore) use the following code

    <br />
12(x-x_0) + 21(y-y_0) -22(z-z_0) = 0

    12(x-x_0) + 21(y-y_0) -22(z-z_0) = 0
    ahhh i see now.

    Thanks so much, your help been amazing useful, learnt what I had no clue about in class
    Last edited by mathsToday; April 6th 2008 at 07:02 AM.
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  12. #12
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    Smile

    So i've got a new equation that needs to be determined by the the fact the following three points lie in it:  (1,2,-4), (2,3,7) and  (4,-1,3) So i find two vectors that lie in the plane.

    v_1= \underbrace{(1,2,-4)}_{Head}-\underbrace{(4,-1,3)}_{tail}= -3 \vec i 3 \vec j -7 \vec k


    and

    v_2= \underbrace{(2,3,7)}_{Head}-\underbrace{(4,-1,3)}_{tail}=-2 \vec i -4 \vec j -4 \vec k

    To find a normal vector we need to take the cross product

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
-3 && 3 && -7 \\<br />
-2 && -4 && -4 \\<br />
\end{vmatrix} = 16 \vec i +2 \vec j- 18 \vec k<br />

    This is the normal vector to the plane

    I just want to check that what i've done is correct so far?
    Last edited by mathsToday; July 26th 2008 at 06:26 AM.
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  13. #13
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    The ideas are correct.
    BUT, you have made many sign errors in almost all the above circulations.
    Go back and check each one again.
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  14. #14
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    Question

    v_1= \underbrace{(1,2,-4)}_{Head}-\underbrace{(4,-1,3)}_{tail}= -3 \vec i, 3 \vec j, -7 \vec k

    and

    v_2= \underbrace{(2,3,7)}_{Head}-\underbrace{(4,-1,3)}_{tail}=-2 \vec i, 4 \vec j , 4 \vec k

    To find a normal vector we need to take the cross product

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
-3 && 3 && -7 \\<br />
-2 && 4 && 4 \\<br />
\end{vmatrix} = 40 \vec i +26 \vec j- 6 \vec k<br />

    This is the normal vector to the plane

    I just want to check that my changes are now correct? thanks.
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  15. #15
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    Quote Originally Posted by TheEmptySet View Post
     12(x-4) + 21(y-1) -22(z-6) = 0

    Distribute out the left hand side.

     12x-48 + 21y-21 -22z+132 = 0  \iff 12x +21y-22z +63=0

    so

    12x+21y-22z=-63
    I really can't remember how we find the answer -63 and can't seem to work it out either can anybody run through the steps to getting the resulting -63?? thanks a lot!
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