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Math Help - Distance between the given parallel planes:

  1. #1
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    Distance between the given parallel planes:

    I had the following problem on my test today and I supposedly got the wrong answer but I can't seem to find where I went wrong. Please let me know if you see an error in my work:

    Find the distance between the given parallel planes: 4z = 6y - 2x (Plane A)
    6z = 2 - 3x + 9y (Plane B)

    Let x = y = 0 on plane 6z = 2 - 3x + 9
    6z = 2
    z = 3
    Thus we have the point (0,0,3) and to find the distance between this point on Plane B and Plane A there is the formula:

    absolute value( a(x0) + b(y0) + c(z0) - D ) / sqrt(a^2 + b^2 + c^2)

    plane A has: a = 2, y = -6, z = 4, d = 0

    so, i did 2(0) - 6(0) + 4(3) = 12 / sqrt( 4 + 36 + 16)

    12 / sqrt(56) and further reduced to 6/sqrt(14)

    the computer assignment says this is wrong... any help is greatly appreciated.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by sj9110 View Post
    I had the following problem on my test today and I supposedly got the wrong answer but I can't seem to find where I went wrong. Please let me know if you see an error in my work:

    Find the distance between the given parallel planes: 4z = 6y - 2x (Plane A)
    6z = 2 - 3x + 9y (Plane B)

    Let x = y = 0 on plane 6z = 2 - 3x + 9
    6z = 2
    z = 3
    Thus we have the point (0,0,3) and to find the distance between this point on Plane B and Plane A there is the formula:

    absolute value( a(x0) + b(y0) + c(z0) - D ) / sqrt(a^2 + b^2 + c^2)

    plane A has: a = 2, y = -6, z = 4, d = 0

    so, i did 2(0) - 6(0) + 4(3) = 12 / sqrt( 4 + 36 + 16)

    12 / sqrt(56) and further reduced to 6/sqrt(14)

    the computer assignment says this is wrong... any help is greatly appreciated.
    I agree with your answer. But try this form: \frac{3\sqrt{14}}{7}
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  3. #3
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    What's the difference? (how did you arrive at that answer?).. also, it was a test so I only had that one shot, what I really want to know is if I have a legitimate claim that I deserve points for this problem so I can bring it up with my professor..
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  4. #4
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    Quote Originally Posted by sj9110 View Post
    What's the difference? (how did you arrive at that answer?).. also, it was a test so I only had that one shot, what I really want to know is if I have a legitimate claim that I deserve points for this problem so I can bring it up with my professor..
    rationalize the den.

    \bigg(\frac{6}{\sqrt{14}}\bigg)\bigg(\frac{\sqrt{1  4}}{\sqrt{14}}\bigg) = \frac{6\sqrt{14}}{14} = \frac{3\sqrt{14}}{7}

    My teacher marks the answer wrong unless it is the simplest form, so I do not think he will give you credit.
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  5. #5
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    Quote Originally Posted by sj9110 View Post
    I had the following problem on my test today and I supposedly got the wrong answer but I can't seem to find where I went wrong. Please let me know if you see an error in my work:
    Sorry, you are not going to get, and should not get, partial credit.

    Find the distance between the given parallel planes: 4z = 6y - 2x (Plane A)
    6z = 2 - 3x + 9y (Plane B)

    Let x = y = 0 on plane 6z = 2 - 3x + 9
    6z = 2
    z = 3
    Here's an obvious error- if 6z= 2, then z= 1/3, not 3!

    Thus we have the point (0,0,3) and to find the distance between this point on Plane B and Plane A there is the formula:

    absolute value( a(x0) + b(y0) + c(z0) - D ) / sqrt(a^2 + b^2 + c^2)

    plane A has: a = 2, y = -6, z = 4, d = 0

    so, i did 2(0) - 6(0) + 4(3) = 12 / sqrt( 4 + 36 + 16)

    12 / sqrt(56) and further reduced to 6/sqrt(14)

    the computer assignment says this is wrong... any help is greatly appreciated.

    Thanks in advance.
    You really shouldn't need to memorize a formula for this- pick any point on the first plane- (0, 0, 0) is an obvious choice. Then construct the line through that point perpendicular to the two planes. The two planes can be written x- 3y+ 2z= 0 and x- 3y+ 2z= 2/3 so a normal line through (0, 0, 0) is x= t, y= -3t, z= 2t. That intersects the second plane when t- 3(-3t)+ 2(2t)= (1+ 9+ 4)t= 14t= 2/3 or t= 1/21. The point of intersection is (1/21, -1/7, 2/21).

    The distance between the planes is the distance from (0, 0, 0) to (1/21, -1/7, 2/21).
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