1. Coordinate Systems

b. Distance Formula - 2D, 3D and nD

2. Distance Formula in 3D

The Pythagorean Theorem can be generalized to \(3\) dimensions and then used to find the distance between two points in \(\mathbb{R}^3\). Consider a rectangular box with sides \(a\), \(b\) and \(c\). The diagonal of the base is \[ r=\sqrt{a^2+b^2} \] So the main diagonal of the box is \[ d=\sqrt{r^2+c^2}=\sqrt{a^2+b^2+c^2} \]

thm_3DPythagBox

Pythagorean Theorem in \(\mathbb{R}^3\)
In \(\mathbb{R}^3\) the main diagonal of a rectangular box with sides \(a, b\), and \(c\) is \[ d=\sqrt{a^2+b^2+c^2} \] which is the square root of the sum of the squares of the sides.

Now consider two points \[ P=(p_1,p_2,p_3) \qquad \text{and} \qquad Q=(q_1,q_2,q_3) \] as shown to the right:

The differences in the \(x,y,\) and \(z\) coordinates are \[ |q_1-p_1|, \quad |q_2-p_2|, \quad \text{and} \quad |q_3-p_3| \] These are the edges of a rectangular box whose main diagonal is the distance between the points \(P\) and \(Q\). Consequently, by the Pythagorean Theorem, we conclude:

def_3DDist

The distance between the points \(P=(p_1,p_2,p_3)\) and \(Q=(q_1,q_2,q_3)\) is: \[ d(P,Q)=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2+(q_3-p_3)^2} \]

Find the distance between the points \(A=(3,1,-4)\) and \(B=(-1,4,8)\)

\(d(A,B)=13\)

\(A=(3,1,-4)\) and \(B=(-1,4,8)\) \[\begin{aligned} d(A,B)&=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2} \\ &=\sqrt{(-1-3)^2+(4-1)^2+(8+4)^2} \\ &=\sqrt{16+9+144}=\sqrt{169}=13 \end{aligned}\]

Equation of a Sphere

By definition, the sphere of radius \(R\) centered at a point \(C=(a,b,c)\) is the set of all points \(X=(x,y,z)\) whose distance from \(C\) is \(R\). Using the distance formula, this sphere is the set of all points \(X=(x,y,z)\) satisfying the equation \[ d(C,X)=R \] or \[ \sqrt{(x-a)^2+(y-b)^2+(z-c)^2}=R. \]

The sphere of radius \(R\) centered at a point \(C=(a,b,c)\) is \[ (x-a)^2+(y-b)^2+(z-c)^2=R^2. \]

Find the equation of the sphere centered at \(P=(2,3,4)\) which is tangent to the \(xz\)-plane

Drop a perpendicular from \(P=(2,3,4)\) to the \(xz\)-plane. This is parallel to the \(y\)-axis. What is the distance from \(P=(2,3,4)\) to the \(xz\)-plane?

\((x-2)^2+(y-3)^2+(z-4)^2=9\)

The distance from \(P=(2,3,4)\) to the \(xz\)-plane is \(3\), the \(y\) coordinate of \(P\). This is the radius. So the sphere is \((x-2)^2+(y-3)^2+(z-4)^2=9\).

Rotate the plot with your mouse.

Some exercises on spheres use the fact that the center of a sphere is at the midpoint of a diameter.

Midpoint of a Line Segment

The midpoint of a line segment is the point halfway between the endpoints. So its coordinates are found by averaging the coordinates of its endpoints. Specifically,

If the endpoints of a line segment are \(P=(a,b,c)\) and \(Q=(d,e,f)\), then the midpoint is: \[ M=\left(\dfrac{a+d}{2}, \dfrac{b+e}{2}, \dfrac{c+f}{2} \right). \]

Find the equation of the sphere which has a diameter with endpoints \(P=(3,-2,7)\) and \(Q=(5,4,-3)\).

The center is at the midpoint of the line segment between \(P=(3,-2,7)\) and \(Q=(5,4,-3)\). The radius is half of the diameter.

\((x-4)^2+(y-1)^2+(z-2)^2=35\)

The center of the sphere is at the midpoint of the line segment between \(P=(3,-2,7)\) and \(Q=(5,4,-3)\) which is found by averaging their coordinates: \[ C=\left(\dfrac{3+5}{2}, \dfrac{-2+4}{2}, \dfrac{7-3}{2}\right)=(4,1,2) \]

The radius of the sphere is half of the distance between \(P\) and \(Q\). This distance is \[\begin{aligned} d(P,Q)&=\sqrt{(5-3)^2+(4--2)^2+(-3-7)^2} \\ &=\sqrt{4+36+100}=2\sqrt{35} \end{aligned}\] and so, the radius is \(r=\sqrt{35}\). So the equation of the sphere is: \[ (x-4)^2+(y-1)^2+(z-2)^2=35 \]

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