# Thread: Distance In 3D

1. ## 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?

2. ## Re: Distance In 3D Originally Posted by harpazo 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.

3. ## Re: Distance In 3D Originally Posted by Walagaster 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.

4. ## Re: Distance In 3D Originally Posted by harpazo 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 $|xz|=\sqrt{105},~\&~|yz|=\sqrt{53}$

5. ## Re: Distance In 3D Originally Posted by Plato 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.