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Thread: Intersection of a plane and a normal.

  1. #1
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    Intersection of a plane and a normal.

    Problem: A normal to $\displaystyle 3x-y+4z=12 $passes through A(5,1,6). Find the point of intersection of the normal and the plane.

    I haven't solved a question like this before in my book, so I am not too sure on how to go about this question. I know that the normal vector to $\displaystyle 3x-y+4z=12 $ is $\displaystyle \vec{n} = (3, -1, 4)$. So, to find the point of intersection, do I take the cross product of the normal and the given point? Or, would I have to find another point on the plane (I'll let $\displaystyle x = 2$ and $\displaystyle y = 1$ to get
    $\displaystyle 3(2) - (1)+4z = 12 => 4z = 7 => z = \frac{7}{4}$, this point will be $\displaystyle B(2, 1,\frac{7}{4})$ and then find $\displaystyle \vec{AB}$ and take the cross product of $\displaystyle \vec{n}$ and $\displaystyle \vec{AB}$?

    I just never encountered a question like this before, so I would appreciate some help on getting started. Thanks in advance.
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  2. #2
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    Find where the line
    $\displaystyle \ell (t) = \left\{ {\begin{array}{*{20}c} {x = 5 + 3t} \\ {y = 1 - t} \\ {z = 6 + 4t} \\ \end{array} } \right.$
    intersects the plane.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Find where the line
    $\displaystyle \ell (t) = \left\{ {\begin{array}{*{20}c} {x = 5 + 3t} \\ {y = 1 - t} \\ {z = 6 + 4t} \\ \end{array} } \right.$
    intersects the plane.
    So, substituting the normal $\displaystyle (3, -1, 4)$:
    $\displaystyle 3 = 5 + 3t => -2 = 3t => x = -\frac{2}{3} $
    $\displaystyle -1 = 1 - t => y = -2$
    $\displaystyle 4 = 6 + 4t => -2 = 4t => z = -\frac{1}{2}$

    Therefore, the line intersects the plane at the point $\displaystyle (-\frac{2}{3}, -2, -\frac{1}{2}$, correct?
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  4. #4
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    No, that's not the point. The normal is a vector, not an (x, y, z) point in the plane. Also note that you have made no use at all of the "= 12" part of the equation of the plane.

    Instead, put x= 5+ 3t, y= 1- t, and z= 6+ 4t in for the x, y, and z in the equation of the plane. Then solve for t and use that to find x, y, and z.
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  5. #5
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    Ah, I see. I'll reattempt this:
    $\displaystyle 3(5 + 3t) - (1-t) + 4 (6 + 4t) = 12$
    $\displaystyle 15 + 9t - 1 + t +24 + 16t = 12 $
    $\displaystyle 26t = -26$
    $\displaystyle t = -1$
    Substituting this back into the original parametric equation:
    $\displaystyle x = 5 + 3(-1) = 2$
    $\displaystyle y = 1 - (-1) = 2$
    $\displaystyle z = 6 + 4(-1) = 2$
    Point of intersection of the normal and the plane is (2, 2, 2). Correct?
    Last edited by Pupil; Jun 4th 2011 at 05:40 PM.
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  6. #6
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    Check it yourself. Does x=2, y= 2, z= 2 satisfy the equation of the plane, 3x- y+ 4z= 12 (so that (2, 2, 2) is on theplane)? Does there exist a single value of t such that x= 5+ 3t= 2, y= 1- t= 2, and z= 6+ 4t= 2 (so that (2, 2, 2) is on the normal line)?
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Check it yourself. Does x=2, y= 2, z= 2 satisfy the equation of the plane, 3x- y+ 4z= 12 (so that (2, 2, 2) is on theplane)? Does there exist a single value of t such that x= 5+ 3t= 2, y= 1- t= 2, and z= 6+ 4t= 2 (so that (2, 2, 2) is on the normal line)?
    Thank you for all the help and effort, HallsofIvy. I now understand perfectly.
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