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Math Help - Planes in 3-space; Describing the equations of a plane

  1. #1
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    Question Planes in 3-space; Describing the equations of a plane

    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: 1, 2, -4), (2, 3, 7), (4, -1, 3)

    So we 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} = 40 \vec i +26 \vec j- 6 \vec k<br />

     40(x-4) + 26(y-1) -6(z-3) = 0

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

     40x+26y-6z = 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!

    So we distribute out the left hand side.

     40x-160 + 26y-26 -6z+18 = 0 ?

    If any mistakes can be pointed out and any help towards the completion of this exercise would be greatly appreciated.
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  2. #2
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    Note that we can also use 20i + 13j - 3k as the normal.
    Then the plane equation is 20x + 13y - 3z = d.
    We can find d simply by using one of the points.
    Check it with another one of the points.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Note that we can also use 20i + 13j - 3k as the normal.
    Then the plane equation is 20x + 13y - 3z = d.
    We can find d simply by using one of the points.
    Check it with another one of the points.
    Thank you for your reply!.
    Ok but the normal i've used is also correct isn't it?, along with my distrubiation of the LHS in the equation:
     40(x-x0) + 26(y-y0) -6(z-z0) = 0
    "We can find simply by using one of the points. Check it with another one of the points."
    That i'm not sure of, how do you mean check it?.
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  4. #4
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    Quote Originally Posted by mathsToday View Post
    Ok but the normal i've used is also correct isn't it?
    The two vectors are parallel. The point is that it is simpler to use small numbers.

    Quote Originally Posted by mathsToday View Post
    "We can find simply by using one of the points. Check it with another one of the points." That i'm not sure of, how do you mean check it?.
    Well use the point  (1,2, - 4):\;20(1) + 13(2) - 3( - 4) = 58. So d=58.
    Check with this point: (4, - 1,3):\;20(4) + 13( - 1) - 3(3) = 58.
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