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Math Help - Reta orthogonal to the plan

  1. #1
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    Reta orthogonal to the plan

    To determine the equations of a line that passes through M r (2.1,-1) and is orthogonal to the plan, ret, and r: \frac{x-2}{3}=\frac{y+1}{2}=z-1 and t: \frac{x}{6}=\frac{y}{4}=\frac{z}{2}


    Answer:

    \frac{x-2}{-3}=y-1=\frac{z+1}{7}
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  2. #2
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    Start with the most general form of a Plane

    Ax + By + Cz + D = 0

    Orthogonal to r

    (1) 3A + 2B + C = 0

    Orthogonal to t

    6A + 4B + 2C = 0 ==> 3A + 2B + C = 0

    Ah! They played a little trick on us. r and t are parallel. Fortunately, we can generate more direction numbers form the two most obvious points on the lines.

    2-0 = 2, -1-0 = -1, and 1-0 = 1

    Orthogonal to a line with those dirction numbers.

    (2) 2A - B + C = 0

    Contains the point (2,1,-1)

    (3) 2A + B - C + D = 0

    That's enough. With three constraints and four variables, we'll have to pick one to parameterize the others. I chose A.

    B = -A/3, C = -7A/3, and D = -4A

    Substituting into the original general form and dividing by -A/3 gives:

    3x - y - 7z - 12 = 0

    There are the values we seek. The line is then trivial.

     <br />
\frac{x-2}{3}\;=\;\frac{y-1}{-1}\;=\;\frac{z+1}{-7}\;\implies\;\frac{x-2}{-3}\;=\;y-1\;=\;\frac{z+1}{7}<br />
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  3. #3
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    Thank you very much. I am Brazilian and I'm trying to communicate better. You have MSN?
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