Distance Formula in Three Dimensions

Derivation of the Distance Formula in 3D

In two-dimensional coordinate geometry, the distance formula is derived using the Pythagorean theorem. The distance formula in three dimensions is a natural extension of this concept, applying the Pythagorean theorem twice in a 3D space. It provides a way to calculate the straight-line distance between any two points in space given their Cartesian coordinates.

Setup

Let $P_1$ and $P_2$ be two points in three-dimensional space with coordinates $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$. Our objective is to find the distance between these two points, denoted as $d$ or $|P_1 P_2|$.

To find this distance, we construct a right-angled triangle in 3D space where the segment $P_1 P_2$ is the hypotenuse. We can do this by considering the changes in the x, y, and z coordinates separately.

Construction and Derivation using Pythagorean Theorem

Consider the points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$. Let's define an intermediate point $Q$ such that it shares the x and y coordinates of $P_2$ but the z-coordinate of $P_1$. The coordinates of $Q$ are $Q(x_2, y_2, z_1)$.

Now consider the line segments $P_1 Q$ and $Q P_2$.

1. Segment $P_1 Q$: This segment connects $P_1(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_1)$. Since the z-coordinates of $P_1$ and $Q$ are the same ($z_1$), the segment $P_1 Q$ is parallel to the XY-plane (specifically, it lies in the plane $z = z_1$). The length of this segment is equivalent to the distance between the points $(x_1, y_1)$ and $(x_2, y_2)$ in a 2D plane. Using the 2D distance formula:

   $|P_1 Q| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

   ... (i)

   Squaring both sides of equation (i) gives:

   $|P_1 Q|^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

   ... (ii)

2. Segment $Q P_2$: This segment connects $Q(x_2, y_2, z_1)$ and $P_2(x_2, y_2, z_2)$. Since the x and y coordinates of $Q$ and $P_2$ are the same ($x_2$ and $y_2$), the segment $Q P_2$ is parallel to the z-axis. The length of this segment is the absolute difference in the z-coordinates:

   $|Q P_2| = |z_2 - z_1|$

   ... (iii)

   Squaring both sides of equation (iii) gives:

   $|Q P_2|^2 = (z_2 - z_1)^2$

   ... (iv)

3. Triangle $P_1 Q P_2$: Now consider the triangle formed by the points $P_1$, $Q$, and $P_2$. The segment $P_1 Q$ lies in the plane $z = z_1$, which is parallel to the XY-plane. The segment $Q P_2$ is parallel to the z-axis, meaning it is perpendicular to any plane parallel to the XY-plane, including the plane $z = z_1$. Therefore, the segment $Q P_2$ is perpendicular to the segment $P_1 Q$. This means $\triangle P_1 Q P_2$ is a right-angled triangle with the right angle at point $Q$. The hypotenuse of this triangle is the segment $P_1 P_2$, whose length is the distance we want to find.

4. Applying the Pythagorean Theorem: In the right-angled triangle $\triangle P_1 Q P_2$, according to the Pythagorean theorem, the square of the hypotenuse ($|P_1 P_2|^2$) is equal to the sum of the squares of the other two sides ($|P_1 Q|^2$ and $|Q P_2|^2$).

   $|P_1 P_2|^2 = |P_1 Q|^2 + |Q P_2|^2$

   (Pythagorean Theorem)

   Substitute the expressions for $|P_1 Q|^2$ from equation (ii) and $|Q P_2|^2$ from equation (iv):

   $|P_1 P_2|^2 = [(x_2 - x_1)^2 + (y_2 - y_1)^2] + (z_2 - z_1)^2$

   ... (v)

   $|P_1 P_2|^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

   ... (vi)

5. Taking the Square Root: To find the distance $|P_1 P_2|$, take the positive square root of both sides of equation (vi):

   $|P_1 P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

   ... (vii)

Equation (vii) is the 3D distance formula.

Distance Formula in 3D

The distance $d$ between any two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ in three-dimensional space is given by:

$\mathbf{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}$

This formula represents the length of the line segment connecting the two points. It is the direct generalization of the 2D distance formula by adding the squared difference of the z-coordinates.

Distance of a Point from the Origin:

A special case of the 3D distance formula is finding the distance of a point $P(x, y, z)$ from the origin $O(0, 0, 0)$. Using the distance formula with $P_1(x_1, y_1, z_1) = O(0, 0, 0)$ and $P_2(x_2, y_2, z_2) = P(x, y, z)$, we get:

$|OP| = \sqrt{(x - 0)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{x^2 + y^2 + z^2}$

So, the distance of a point $(x, y, z)$ from the origin is $\mathbf{\sqrt{x^2 + y^2 + z^2}}$.

Calculating Distance Between Two Points in 3D Space

The 3D distance formula allows us to find the length of the straight line segment connecting any two points in three-dimensional space, provided we know their Cartesian coordinates. The formula is a direct generalization of the 2D distance formula.

Given two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$, the distance $d$ between them is calculated using the formula:

$\mathbf{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}$

Alternatively, the formula can be written as $d = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$, where $\Delta x = x_2 - x_1$, $\Delta y = y_2 - y_1$, and $\Delta z = z_2 - z_1$ are the differences in the corresponding coordinates.

Steps to Calculate Distance in 3D:

To find the distance between two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$:

1. Identify Coordinates: Clearly note the coordinates of the two points. Let the first point be $(x_1, y_1, z_1)$ and the second point be $(x_2, y_2, z_2)$. The order of points $P_1$ and $P_2$ does not affect the final distance, as squaring the coordinate differences makes them non-negative.

2. Calculate Coordinate Differences: Subtract the corresponding coordinates of the first point from the second point. Calculate the difference in x-coordinates ($x_2 - x_1$), the difference in y-coordinates ($y_2 - y_1$), and the difference in z-coordinates ($z_2 - z_1$).

3. Square the Differences: Square each of the three differences obtained in Step 2: $(x_2 - x_1)^2$, $(y_2 - y_1)^2$, and $(z_2 - z_1)^2$. Remember that squaring a negative number results in a positive number.

4. Sum the Squared Differences: Add the three squared differences obtained in Step 3: $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$. The sum will always be non-negative.

5. Take the Square Root: Calculate the positive square root of the sum obtained in Step 4. This final value is the distance $d$ between the two points.

   $d = \sqrt{\text{Sum from Step 4}}$

Answer:

Answer:

Applications in 3D Geometry

The 3D distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$, is a fundamental tool for solving various geometric problems in three-dimensional space. By calculating the lengths of line segments between points, we can verify geometric properties, classify figures, and solve problems involving distances.

1. Checking Collinearity of Three Points

Three points A, B, and C are collinear (lie on the same straight line) if and only if the sum of the lengths of any two of the segments formed by these points is equal to the length of the third segment. For example, if B lies between A and C on the line, then the distance from A to B plus the distance from B to C must equal the distance from A to C: $|AB| + |BC| = |AC|$.

To check if three points A, B, and C are collinear using the 3D distance formula:

1. Calculate the distances between all three pairs of points: $|AB|$, $|BC|$, and $|AC|$ using the 3D distance formula.
2. Check if the sum of the lengths of the two shorter segments equals the length of the longest segment. If this condition is met for any permutation of the points, they are collinear. Otherwise, they form a triangle.

Answer:

2. Identifying Types of Triangles

Given the coordinates of the three vertices of a triangle, say A, B, and C, we can determine the type of triangle by calculating the lengths of its sides $|AB|$, $|BC|$, and $|CA|$ using the 3D distance formula. Then, we compare these lengths and their squares.

• Scalene Triangle: All three sides have different lengths ($|AB| \neq |BC|$, $|BC| \neq |CA|$, and $|CA| \neq |AB|$).
• Isosceles Triangle: At least two sides have equal lengths (e.g., $|AB| = |BC|$).
• Equilateral Triangle: All three sides have equal lengths ($|AB| = |BC| = |CA|$).
• Right-Angled Triangle: The triangle is right-angled if the converse of the Pythagorean theorem holds true. This means the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides (e.g., if $|AC|$ is the longest side, check if $|AB|^2 + |BC|^2 = |AC|^2$).

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3. Identifying Types of Quadrilaterals

Given the coordinates of four vertices A, B, C, and D, we can classify the quadrilateral formed by these points by calculating the lengths of all four sides ($|AB|, |BC|, |CD|, |DA|$) and the lengths of the two diagonals ($|AC|, |BD|$).

• Parallelogram: Opposite sides are equal in length ($|AB| = |CD|$ and $|BC| = |DA|$). The diagonals generally have different lengths.
• Rectangle: A parallelogram with equal diagonals. So, opposite sides are equal ($|AB| = |CD|$ and $|BC| = |DA|$), and the diagonals are equal ($|AC| = |BD|$). Alternatively, verify opposite sides are equal and check if the Pythagorean theorem holds for one of the corner triangles (e.g., $|AB|^2 + |BC|^2 = |AC|^2$).
• Rhombus: All four sides are equal in length ($|AB| = |BC| = |CD| = |DA|$). The diagonals generally have different lengths and are perpendicular bisectors of each other (perpendicularity checked using slopes/vectors).
• Square: A rhombus with equal diagonals. So, all four sides are equal ($|AB| = |BC| = |CD| = |DA|$), and the diagonals are equal ($|AC| = |BD|$). This implies it is both a rectangle and a rhombus.

4. Finding Points Equidistant from Given Points

The distance formula is used to set up equations when a point is equidistant from other given points. This can lead to finding the equation of a locus or the coordinates of a specific point satisfying the condition.

• The locus of a point in 3D space equidistant from two fixed points P and Q is the perpendicular bisector plane of the segment PQ. If $A(x,y,z)$ is any such point, then $|AP| = |AQ|$, which leads to $|AP|^2 = |AQ|^2$. Expanding this using the distance formula gives a linear equation in x, y, and z, representing a plane.
• You might be asked to find a point on a specific axis or plane that is equidistant from given points.

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