# 1. Coordinate Systems

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

## 3. Distance Formula in Higher Dimensions (Optional)

The Pythagorean Theorem can be generalized to \(n\) dimensions and then used to find the distance between two points in \(\mathbb{R}^n\). Consider a \(n\) dimensional rectangular box with sides \(a_1,a_2,\cdots,a_n\).

In \(\mathbb{R}^n\) the main diagonal of a rectangular box with sides \(a_1, a_2,\cdots,a_n\) is \[ d=\sqrt{a_1^2+a_2^2+\cdots+a_n^2} \] which is the square root of the sum of the squares of the sides.

Now consider two points \(P=(p_1,p_2,\cdots,p_3)\) and \(Q=(q_1,q_2,\cdots,q_3)\). The changes in the coordinates are \(|q_1-p_1|,|q_2-p_2|,\cdots,|q_n-p_n|\). These are the edges of an \(n\) dimensional rectangular box whose main diagonal is the distance between the points \(P\) and \(Q\).

By the Pythagorean Theorem: \[ d(P,Q)=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2+\cdots+(q_n-p_n)^2} \]

Find the distance between the points \(A=(6,1,3,5)\) and \(B=(2,3,7,-3)\) in \(\mathbb{R}^4\)

\(d(A,B)=10\)

\(A=(6,1,3,5)\) and \(B=(2,3,7,-3)\) \[\begin{aligned} d(A,B)&=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2+(b_4-a_4)^2} \\ &=\sqrt{(2-6)^2+(3-1)^2+(7-3)^2+(-3-5)^2} \\ &=\sqrt{16+4+16+64}=\sqrt{100}=10 \end{aligned}\]

### Equation of an \(n\)-Sphere in \(\mathbb{R}^n\)

By definition, the \(n\)-sphere of radius \(R\) centered at a point \(C=(a_1,a_2,\cdots,a_n)\) is the set of all points \(X=(x_1,x_2,\cdots,x_n)\) whose distance from \(C\) is \(R\). Using the distance formula, the \(n\)-sphere is the set of all points \(X=(x_1,x_2,\cdots,x_n)\) satisfying the equation \[ d(C,X)=R \] or \[ \sqrt{(x_1-a_1)^2+(x_2-a_2)^2+\cdots+(x_n-a_n)^2}=R. \]

The sphere of radius \(R\) centered at a point \(C=(a_1,a_2,\cdots,a_n)\) is \[ (x_1-a_1)^2+(x_2-a_2)^2+\cdots+(x_n-a_n)^2=R^2. \]

Find the equation of the \(4\)-sphere for which \(A=(6,1,3,5)\) and \(B=(2,3,7,-3)\) are endpoints of a diagonal. Take the coordinates to be \((w,x,y,z)\).

The diameter is the distance from \(A\) to \(B\).

The center is the midpoint of the segment from \(A\) to \(B\). This is found by averaging the coordinates of \(A\) and \(B\), i.e. \(C=\dfrac{A+B}{2}\).

\((w-4)^2+(x-2)^2+(y-5)^2+(z-1)^2=25\)

The distance from \(A=(6,1,3,5)\) to \(B=(2,3,7,-3)\). was found in the previous exercise to be \(10\). So the radius is \(5\).

The center is the average of \(A\) to \(B\): \[ C=\dfrac{A+B}{2} =\left(\dfrac{6+2}{2},\dfrac{1+3}{2},\dfrac{3+7}{2},\dfrac{5-3}{2}\right) =(4,2,5,1) \] So the equation of the \(4\)-sphere is \[ (w-4)^2+(x-2)^2+(y-5)^2+(z-1)^2=25 \]

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