Chapter 12 Introduction to Three Dimensional Geometry (Concepts)

Cartesian Coordinate System in Three Dimensions

Until now, our study of analytical geometry has been limited to a two-dimensional plane. However, we live in a physical world that is three-dimensional. To describe the position of objects in space, we must extend our mathematical knowledge from two dimensions (represented by an ordered pair) to three-dimensional space (represented by an ordered triplet).

Cartesian Coordinate System

The study of three-dimensional geometry is a natural extension of two-dimensional Cartesian geometry. In a plane, we require two numbers to locate a point, whereas, in space, we require three independent real numbers. This system is essential for fields ranging from Aerospace Engineering in organizations like ISRO to Quantum Physics.

The Rectangular Coordinate Axes

To identify the position of a point in space, we consider three mutually perpendicular straight lines passing through a fixed point $O$ (the Origin). These lines are the Rectangular Coordinate Axes.

$\bullet$ X-axis ($X'OX$): The horizontal line representing the first dimension.

$\bullet$ Y-axis ($Y'OY$): The line perpendicular to the $x$-axis in the horizontal plane.

$\bullet$ Z-axis ($Z'OZ$): The vertical line perpendicular to both $x$ and $y$ axes.

Right-Handed Coordinate System: If you curl the fingers of your right hand from the positive $x$-axis toward the positive $y$-axis, your thumb points toward the positive $z$-axis.

The Coordinate Planes

The three axes taken in pairs determine three fundamental planes called Coordinate Planes. Every point in space lies either on these planes or in the regions between them.

1. The XY-Plane

This plane contains the $x$ and $y$ axes. For any point in this plane, the vertical distance (height) is zero. Thus, its mathematical equation is:

2. The YZ-Plane

This plane contains the $y$ and $z$ axes. For any point here, the distance along the $x$-direction is zero.

3. The ZX-Plane

This plane contains the $z$ and $x$ axes. Here, the $y$-coordinate is always zero.

The intersection of these three planes is the Origin $O(0, 0, 0)$. The intersection of the XY-plane and YZ-plane is the Y-axis, and so on.

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Coordinates of a Point in Space

To locate a point in a three-dimensional space, we need three independent measurements. This is a direct extension of the two-dimensional system. Every point in space corresponds to a unique ordered triplet of real numbers $(x, y, z)$, and conversely, every ordered triplet $(x, y, z)$ corresponds to a unique point in space. This is known as a one-one correspondence between the set of points in space and the set of ordered triplets.

Method 1: The Parallelopiped Method

This method is purely geometric and relies on the construction of a rectangular box (parallelopiped). Imagine a point $P$ suspended in a room. To find its "address":

1. Through point $P$, we pass three planes. Each plane is constructed to be parallel to one of the coordinate planes ($XY$, $YZ$, and $ZX$).

2. These planes will intersect the three coordinate axes ($x, y, z$) at points $A, B$, and $C$ respectively.

3. The origin $O$, the points on the axes $A, B, C$, and the point $P$ (along with three other intersection points) form the vertices of a rectangular parallelopiped.

The lengths of the edges of this box starting from the origin represent the coordinates:

Method 2: The Perpendicular Distance Method

This method is more "procedural" and is often used in vector analysis and physics. It involves projecting the point onto a plane and then onto an axis.

Step 1: Projection onto the Plane

Drop a perpendicular from point $P$ to the $XY$-plane. Let the foot of this perpendicular be $M$. The length of this perpendicular line segment $PM$ gives the $z$-coordinate. It represents the "height" of the point above (or below) the floor ($XY$-plane).

Step 2: Projection onto the Axis

Now, from the point $M$ (which is in the $XY$-plane), drop another perpendicular to the $x$-axis. Let the foot of this perpendicular be $N$.

$\bullet$ The distance $ON$ along the $x$-axis gives the $x$-coordinate.

$\bullet$ The distance $NM$ (which is parallel to the $y$-axis) gives the $y$-coordinate.

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Important Geometric Observations

In three-dimensional analytical geometry, the coordinates of a point are not just numbers; they represent specific geometric relationships with respect to the Origin, Axes, and Coordinate Planes.

1. The Origin and Directed Distances

The origin $O(0, 0, 0)$ is the unique point where all three coordinate axes intersect. Every coordinate of a point $P(x, y, z)$ can be interpreted as a directed distance from a specific coordinate plane.

Physical Interpretation of Coordinates:

$\bullet$ The $x$-coordinate is the directed distance of point $P$ from the YZ-plane ($x = 0$).

$\bullet$ The $y$-coordinate is the directed distance of point $P$ from the ZX-plane ($y = 0$).

$\bullet$ The $z$-coordinate is the directed distance of point $P$ from the XY-plane ($z = 0$).

If a coordinate is positive, the point is on the positive side of the respective plane; if negative, it is on the opposite side.

2. Analysis of Points on Coordinate Planes

A coordinate plane is a flat surface formed by any two axes. When a point lies on such a plane, its distance from that plane is zero, leading to the following properties:

Coordinate Plane	Fixed Property	General Point Format	Equation of Plane
XY-plane	Distance from $XY$ is $0$	$(x, y, 0)$	$z = 0$
YZ-plane	Distance from $YZ$ is $0$	$(0, y, z)$	$x = 0$
ZX-plane	Distance from $ZX$ is $0$	$(x, 0, z)$	$y = 0$

3. Analysis of Points on Coordinate Axes

When a point lies on a specific axis, it effectively lies at the intersection of two coordinate planes. Consequently, two of its coordinates must be zero.

General Format of Points on Axes:

Axis Location	On X-axis	On Y-axis	On Z-axis
General Point	$(x, 0, 0)$	$(0, y, 0)$	$(0, 0, z)$
Equations	$y = 0, z = 0$	$x = 0, z = 0$	$x = 0, y = 0$

4. Concepts of Projection and Symmetry (Symmetry/Mirror Image)

Foot of Perpendicular:

The foot of the perpendicular from $P(x, y, z)$ to a plane is the point on that plane where the specific coordinate becomes zero. For example, the foot of perpendicular on the $YZ$-plane is $(0, y, z)$.

Mirror Image of a Point:

The image of a point $P(x, y, z)$ with respect to a plane or origin is found by changing the sign of the coordinate perpendicular to the plane of reflection.

Reflection Relative to	Image Coordinates
XY-plane	$(x, y, -z)$
YZ-plane	$(-x, y, z)$
ZX-plane	$(x, -y, z)$
Origin	$(-x, -y, -z)$

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Signs of Coordinates in Octants

In two-dimensional geometry, the $x$ and $y$ axes divide the plane into four quadrants. In three-dimensional geometry, the three mutually perpendicular coordinate planes ($XY$, $YZ$, and $ZX$ planes) divide the entire space into eight compartments, known as octants.

Naming Convention of Octants

The octants are named based on the regions bounded by the positive and negative directions of the axes starting from the origin $O$. The notation used is as follows:

1. Octant I: $OXYZ$ (All coordinates are positive)

2. Octant II: $OX'YZ$ ($x$ is negative, $y$ and $z$ are positive)

3. Octant III: $OX'Y'Z$ ($x$ and $y$ are negative, $z$ is positive)

4. Octant IV: $OXY'Z$ ($y$ is negative, $x$ and $z$ are positive)

5. Octant V: $OXYZ'$ ($z$ is negative, $x$ and $y$ are positive)

6. Octant VI: $OX'YZ'$ ($x$ and $z$ are negative, $y$ is positive)

7. Octant VII: $OX'Y'Z'$ (All coordinates are negative)

8. Octant VIII: $OXY'Z'$ ($y$ and $z$ are negative, $x$ is positive)

Sign Determination Table

The following table summarizes the signs of the coordinates for any point $P(x, y, z)$ based on the octant it occupies:

Octants →	I	II	III	IV	V	VI	VII	VIII
$x$	$+$	$-$	$-$	$+$	$+$	$-$	$-$	$+$
$y$	$+$	$+$	$-$	$-$	$+$	$+$	$-$	$-$
$z$	$+$	$+$	$+$	$+$	$-$	$-$	$-$	$-$

Practical Rule for Identification

A quick way to identify the octant of a point $(x, y, z)$ is to follow these steps:

Step 1: Look at the $z$-coordinate. If $z$ is positive, the point lies in one of the first four octants (I, II, III, or IV). If $z$ is negative, it lies in one of the last four octants (V, VI, VII, or VIII).

Step 2: Determine the 2D quadrant based on $(x, y)$. If $z$ is positive, the octant number is the same as the quadrant number. If $z$ is negative, add 4 to the quadrant number of $(x, y)$.

For example, if $(x, y)$ is in the 3rd quadrant (both negative) and $z$ is negative, the octant is $3 + 4 = 7$ (Octant VII).

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Distance Formula in Three Dimensions

To determine the distance between any two points in three-dimensional space, we extend the concept of the 2D distance formula by incorporating the $z$-coordinate. This is achieved through the application of the Pythagoras Theorem in a rectangular parallelopiped.

Derivation of the Distance Formula

Given:

Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be the two given points in space.

To Find:

The length of the line segment $PQ$.

Construction Required:

1. Through points $P$ and $Q$, draw planes parallel to the coordinate planes to form a rectangular parallelopiped (box) where $PQ$ is the body diagonal.

2. Let $A$ and $M$ be vertices of this box such that $PA$ is an edge parallel to the $x$-axis, $AM$ is an edge parallel to the $y$-axis, and $MQ$ is an edge parallel to the $z$-axis.

3. Join $PM$ to form the face diagonal of the base $PAMB$.

Proof:

In the rectangular box, $MQ$ is perpendicular to the plane $PAMB$. Since $PM$ lies in this plane, $MQ \perp PM$. Therefore, $\angle PMQ = 90^\circ$.

In $\triangle PMQ$, by applying Pythagoras Theorem:

Now, consider the base of the box. Since $AM$ is perpendicular to $AP$ (as they are parallel to $y$ and $x$ axes respectively), $\angle MAP = 90^\circ$.

In $\triangle AMP$, by applying Pythagoras Theorem:

Substituting the value of $PM^2$ from equation (ii) into equation (i), we get:

From the coordinates of $P$ and $Q$, the lengths of the edges are:

$AP = x_2 - x_1$

$AM = y_2 - y_1$

$MQ = z_2 - z_1$

Substituting these values into equation (iii):

$PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

Taking the square root of both sides:

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

Special Cases

1. Distance from the Origin

The distance of any point $P(x_1, y_1, z_1)$ from the origin $O(0, 0, 0)$ is calculated by substituting $x_2=0, y_2=0, z_2=0$ into the formula:

$OP = \sqrt{(x_1 - 0)^2 + (y_1 - 0)^2 + (z_1 - 0)^2}$

$OP = \sqrt{x_1^2 + y_1^2 + z_1^2}$

2. Locus of a Point

The locus of a point in space is the surface or path formed by a moving point under specific geometric conditions. For example, if a point moves such that its distance from a fixed point is constant, its locus is a sphere.

Common 3D Loci Examples:

1. Sphere: The locus of a point $P(x, y, z)$ which moves such that its distance from a fixed point $C(h, k, l)$ is always a constant $r$. Using the distance formula, the equation of the sphere is:

$\sqrt{(x - h)^2 + (y - k)^2 + (z - l)^2} = r$

Squaring both sides:

$(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$

2. Perpendicular Bisector Plane: The locus of a point $P(x, y, z)$ which is equidistant from two fixed points $A$ and $B$. The condition is $PA = PB$.

3. Cylinder: The locus of a point which remains at a constant distance from a fixed line (the axis of the cylinder).

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Section Formula in Three Dimensions

The Section Formula is used to determine the coordinates of a point which divides a line segment joining two given points in a specific ratio, either internally or externally. This is a vital tool in 3D analytical geometry for finding positions along a path or locating specific geometric centers.

Derivation of the Internal Section Formula

Given: Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be two given points in space. Let $R(x, y, z)$ be a point on the line segment $PQ$ such that it divides $PQ$ internally in the ratio $m_1 : m_2$.

To Find: The coordinates $(x, y, z)$ of the point $R$.

Construction Required:

1. Draw perpendiculars $PM, QN$, and $RL$ to the $XY$-plane from points $P, Q$, and $R$ respectively.

2. The points $M, N,$ and $L$ are the feet of these perpendiculars and lie on a straight line in the $XY$-plane.

3. Through $R$, draw a straight line $ARB$ parallel to $MLN$ to intersect the line $MP$ (produced) at $A$ and the line $NQ$ at $B$.

Mathematical Proof

The lines $PM, RL,$ and $QN$ are parallel as they are all perpendicular to the same $XY$-plane. Since they are intersected by the transversal $PQ$, the points $P, R,$ and $Q$ are coplanar.

In $\triangle APR$ and $\triangle BQR$:

1. $\angle PAR = \angle QBR = 90^\circ$ (By construction of parallel lines and perpendiculars).

2. $\angle PRA = \angle QRB$ (Vertically opposite angles).

Therefore, by $AA$ similarity criterion, $\triangle APR \sim \triangle BQR$.

From the property of similar triangles, the ratios of the corresponding sides are equal:

From the diagram, the directed distances parallel to the $z$-axis are:

$PA = RL - PM = z - z_1$

$BQ = QN - RL = z_2 - z$

Substituting these values into equation (i):

On cross-multiplying, we get:

Similarly, by projecting the points onto the $YZ$ and $ZX$ planes, we can derive the $x$ and $y$ coordinates:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Summary Table for Internal Division

The coordinates of the point $R$ dividing $PQ$ in the ratio $m_1 : m_2$ are summarized below:

Coordinate	Internal Formula
x-coordinate	$\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$
y-coordinate	$\frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$
z-coordinate	$\frac{m_1 z_2 + m_2 z_1}{m_1 + m_2}$

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External Division in Three Dimensions

When a point $R$ lies on the line passing through $P$ and $Q$ but is located outside the segment $PQ$, it is said to divide the segment $PQ$ externally in the ratio $m_1 : m_2$. This means the point $R$ is located on the line segment $PQ$ produced.

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Conceptual Understanding

In external division, the ratio is defined as the directed distance from the first point to the dividing point, compared to the distance from the dividing point to the second point. Mathematically:

Unlike internal division, where $R$ is between $P$ and $Q$, in external division, $R$ can be situated in two possible positions:

1. If $m_1 > m_2$: The point $R$ lies on the side of $Q$ (i.e., $PQ$ is produced towards $Q$).

2. If $m_1 < m_2$: The point $R$ lies on the side of $P$ (i.e., $QP$ is produced towards $P$).

Comparison of Internal and External Division

The following diagrams illustrate the geometric difference between the two types of division:

$\bullet$ Internal Division: The segments $PR$ and $RQ$ have the same direction. The ratio $m_1 : m_2$ is considered positive.

$\bullet$ External Division: The segments $PR$ and $RQ$ have opposite directions. In mathematical calculations, this is represented by taking the ratio $m_1 : m_2$ as negative.

Derivation of External Section Formula

The External Section Formula determines the coordinates of a point $R$ that lies on the extension of the line segment joining two points $P$ and $Q$. In this case, the point $R$ is outside the segment $PQ$, and the ratio is measured from the extremities to this external point.

Given:

Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be two given points. Let $R(x, y, z)$ be the point that divides the line segment $PQ$ externally in the ratio $m_1 : m_2$. This implies:

Construction Required:

1. Draw perpendiculars $PM, QN,$ and $RL$ from the points $P, Q,$ and $R$ respectively to the $XY$-plane.

2. The feet of these perpendiculars $M, N,$ and $L$ lie on a straight line in the $XY$-plane.

3. Through $P$, draw a line parallel to $MLN$ to meet the perpendiculars $QN$ and $RL$ (produced if necessary) at points $A$ and $B$ respectively.

Proof:

Since $PM, QN,$ and $RL$ are perpendicular to the same plane, they are parallel to each other. The triangles $\triangle PAQ$ and $\triangle PBR$ are similar by the $AA$ (Angle-Angle) similarity criterion, as they share a common angle at $P$ and have corresponding right angles.

However, to stay consistent with the ratio $\frac{m_1}{m_2}$ defined as $\frac{PR}{RQ}$, we consider the similarity of triangles formed between the segments. By the property of parallel lines and transversals, the ratio of segments on the line $PQ$ is equal to the ratio of the segments of the perpendiculars (the $z$-distances).

From the similarity of triangles, we have:

Substituting the $z$-coordinates for the lengths of the perpendiculars:

$\bullet$ $RL = z$

$\bullet$ $PM = z_1$

$\bullet$ $QN = z_2$

Substituting these into equation (ii) along with the given ratio $m_1/m_2$:

Now, we solve for $z$ through cross-multiplication:

$m_1(z - z_2) = m_2(z - z_1)$

$m_1 z - m_1 z_2 = m_2 z - m_2 z_1$

Rearranging the terms to group $z$ on one side:

$m_1 z - m_2 z = m_1 z_2 - m_2 z_1$

$z(m_1 - m_2) = m_1 z_2 - m_2 z_1$

By following the same procedure and projecting the line segment onto the $YZ$-plane and $ZX$-plane, we can derive the $x$ and $y$ coordinates:

$x = \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}$

$y = \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2}$

Thus, the coordinates of the point $R$ dividing $PQ$ externally in the ratio $m_1 : m_2$ are:

$R \left( \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}, \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2}, \frac{m_1 z_2 - m_2 z_1}{m_1 - m_2} \right)$

The $k : 1$ Ratio Method

When we are required to find the ratio in which a point divides a line segment, using two unknowns $m_1$ and $m_2$ makes the calculation cumbersome. Instead, we divide both the numerator and denominator of the section formula by $m_2$ and let $k = \frac{m_1}{m_2}$.

Derivation of the $k : 1$ Coordinates

Starting with the standard internal division formula for the $x$-coordinate:

Dividing both the numerator and denominator by $m_2$:

By substituting $k = \frac{m_1}{m_2}$, the coordinates of the dividing point $R(x, y, z)$ become:

Significance of the Sign of $k$

The value of $k$ obtained after solving an equation provides immediate information about the position of the point $R$ relative to the segment $PQ$:

1. Internal Division: If $k$ is positive ($k > 0$), it implies $m_1$ and $m_2$ have the same sign. The point $R$ lies between $P$ and $Q$.

2. External Division: If $k$ is negative ($k < 0$), it implies $m_1$ and $m_2$ have opposite signs. The point $R$ lies on the line $PQ$ produced.

The Mid-point Formula

The mid-point is a unique case where the point $R$ divides the segment $PQ$ into two equal halves. Thus, the ratio is $1 : 1$, meaning $m_1 = 1$ and $m_2 = 1$ (or $k = 1$).

Substituting these values into the section formula:

Similarly for $y$ and $z$ coordinates, the Mid-point of segment $PQ$ is given by:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$

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Centroid of a Triangle in Three Dimensions

The centroid of a triangle is the point where the three medians of the triangle intersect. A median is a line segment joining a vertex of the triangle to the mid-point of the opposite side. In three-dimensional geometry, the centroid represents the geometric centre of the triangular region.

Properties of the Centroid

1. Concurrency: All three medians of a triangle are concurrent, meaning they meet at a single point (the centroid).

2. Ratio: The centroid ($G$) divides each median in the ratio $2 : 1$, measuring from the vertex to the mid-point of the opposite side.

Derivation of the Centroid Formula

Given:

Let $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$ be the vertices of $\triangle ABC$. Let $G(x, y, z)$ be the centroid of the triangle.

Step 1: Finding the Mid-point of the Base

Let $D$ be the mid-point of the side $BC$. By the mid-point formula, the coordinates of $D$ are:

Step 2: Applying the Section Formula

The centroid $G$ lies on the median $AD$ and divides it in the ratio $m_1 : m_2 = 2 : 1$ internally.

Using the internal section formula for the $x$-coordinate of $G$:

$x = \frac{m_1x_D + m_2x_A}{m_1 + m_2}$

Substituting $m_1 = 2, m_2 = 1, x_D = \frac{x_2 + x_3}{2}$ and $x_A = x_1$:

$x = \frac{2 \left( \frac{x_2 + x_3}{2} \right) + 1(x_1)}{2 + 1}$

$x = \frac{x_2 + x_3 + x_1}{3}$

Similarly, for the $y$ and $z$ coordinates:

$y = \frac{y_1 + y_2 + y_3}{3}$

$z = \frac{z_1 + z_2 + z_3}{3}$

Final Centroid Formula:

Remark on Symmetry

The coordinates of $G$ are symmetric with respect to the coordinates of the vertices $A, B,$ and $C$. This symmetry mathematically proves that whichever median is chosen ($AD, BE,$ or $CF$), the resulting coordinates for the point of division $2:1$ will always be the same. This confirms that the medians are concurrent at $G$.

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