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Thread: Distance In 3D

  1. #1
    Super Member harpazo's Avatar
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    Distance In 3D

    Given x = (-1, 4), y = (6, 2), and z = (4, -5), find the distance from x to y, y to z, and x to z. State whether the triangle is an
    isosceles, right, neither, or both.

    Let me see what is being asked here.

    1. I must use the distance formula for points of the type
    (x, y, z) three times to find the distance from x to y, y to z, and x to z, respectively.

    2. I must then say if the triangle formed by the distances found in part 1 above describe an isosceles triangle, a right triangle, neither triangles or both triangles.

    Is this right?
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  2. #2
    Member Walagaster's Avatar
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    Re: Distance In 3D

    Quote Originally Posted by harpazo View Post
    Given x = (-1, 4), y = (6, 2), and z = (4, -5), find the distance from x to y, y to z, and x to z. State whether the triangle is an
    isosceles, right, neither, or both.

    Let me see what is being asked here.

    1. I must use the distance formula for points of the type
    (x, y, z) three times to find the distance from x to y, y to z, and x to z, respectively.

    2. I must then say if the triangle formed by the distances found in part 1 above describe an isosceles triangle, a right triangle, neither triangles or both triangles.

    Is this right?
    This is not a 3D problem because you points are all in 2D. And it is poor notation to use x and y as the names of points and also as coordinates like in (x,y). But. yes, you can use the 2D distance formula to solve the problem.
    Thanks from topsquark and harpazo
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  3. #3
    Super Member harpazo's Avatar
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    Re: Distance In 3D

    Quote Originally Posted by Walagaster View Post
    This is not a 3D problem because you points are all in 2D. And it is poor notation to use x and y as the names of points and also as coordinates like in (x,y). But. yes, you can use the 2D distance formula to solve the problem.
    There is a typo at my end. I will check the book tomorrow and make all corrections.
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    Re: Distance In 3D

    Quote Originally Posted by harpazo View Post
    Given x = (-1, 4), y = (6, 2), and z = (4, -5), find the distance from x to y, y to z, and x to z. State whether the triangle is an
    isosceles, right, neither, or both.

    Let me see what is being asked here.

    1. I must use the distance formula for points of the type
    (x, y, z) three times to find the distance from x to y, y to z, and x to z, respectively.

    2. I must then say if the triangle formed by the distances found in part 1 above describe an isosceles triangle, a right triangle, neither triangles or both triangles.
    You have been told that this is not in $\mathbb{R}^3$ Now let $|xy|$ be the distance from $x$ to $y$. So $|xy|=\sqrt{(-1-6)^2+(4-2)^2}=\sqrt{49+4}$
    Hence $|xz|=\sqrt{105},~\&~|yz|=\sqrt{53}$
    Thanks from harpazo
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  5. #5
    Super Member harpazo's Avatar
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    Re: Distance In 3D

    Quote Originally Posted by Plato View Post
    You have been told that this is not in $\mathbb{R}^3$ Now let $|xy|$ be the distance from $x$ to $y$. So $|xy|=\sqrt{(-1-6)^2+(4-2)^2}=\sqrt{49+4}$
    Hence $|xz|=\sqrt{105},~\&~|yz|=\sqrt{53}$
    I made a typo. This is not in 3D or R^3. I know how to solve this problem after looking at the question a bit more carefully.
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