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  1. #1
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    distance question

    The point (3,2,-1) lies equidistant from two parallel planes Psub1 and Psub2.
    Psub1 has equation 2x-y+2z+4=0.
    Find an equation for Psub2.

    Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    The point (3,2,-1) lies equidistant from two parallel planes Psub1 and Psub2.
    Psub1 has equation 2x-y+2z+4=0.
    Find an equation for Psub2.

    Thank you very much.
    Hint, a parallel plane has equation,
    2x-y+2z+K=0
    All you need to do if find K.
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  3. #3
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    Do you know the formula for the distance between a point and a plane?.

    The plane has equation ax+by+cz+d=0

    The point: (x_{0}, y_{0}, z_{0})

    D=\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\sqrt{a^{2}+b^{2  }+c^{2}}}

    You have:

    \frac{|2(3)+(-1)(2)+2(-1)+4|}{\sqrt{2^{2}+(-1)^{2}+2^{2}}}=2

    Now, use the formula again and solve for k as PH said:

    \frac{|2(3)+(-1)(2)+2(-1)+k|}{\sqrt{2^{2}+(-1)^{2}+2^{2}}}=2
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  4. #4
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    Hi galactus,

    solving what you gave : the result of k will be 4 again!

    So how can we get the equation of Psub2?
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  5. #5
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    No, one of the answers is 4. There's another.
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  6. #6
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    Hi galactus,

    Could you please show me how? Thank you very much.
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  7. #7
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    Jenny20, I am sorry, but I gave you the entire equation. Solve for k.

    I don't what to seem 'short', but if you can't solve that, you shouldn't be in calc.

    What it amounts to is solving \frac{|6-2-2+k|}{3}=2

    4 is one of the answers. What's the other?.
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