Coordinate Axes and Coordinate Planes in Three Dimensional Space
To describe the position of a point in space in a precise and systematic manner, mathematics uses a three-dimensional coordinate system. This system is formed by three mutually perpendicular straight lines, called the coordinate axes, which intersect at a common point known as the origin. These axes are named the \(x\)-axis, \(y\)-axis, and \(z\)-axis, and they extend indefinitely in both positive and negative directions.
Each pair of coordinate axes determines a flat surface called a coordinate plane. The plane formed by the \(x\)-axis and the \(y\)-axis is known as the \(xy\)-plane, the plane formed by the \(y\)-axis and the \(z\)-axis is called the \(yz\)-plane, and the plane formed by the \(z\)-axis and the \(x\)-axis is called the \(zx\)-plane. These three coordinate planes divide the entire space into eight distinct regions, called octants.
Any point in three-dimensional space can be uniquely represented by an ordered triplet \(x,\,y,\,z\), where the numbers \(x,\,y\) and zdenote the directed distances of the point from the \(yz\)-plane, \(zx\)-plane, and \(xy\)-plane respectively, measured parallel to the corresponding coordinate axes. Thus, the coordinate axes and coordinate planes together provide a fundamental framework for locating points, studying geometrical figures, and developing algebraic representations of spatial relationships in three-dimensional geometry.
Coordinates of a Point in Space
In three-dimensional geometry, the position of a point in space is described with the help of a system of three mutually perpendicular coordinate axes. When a point is fixed in space, its location cannot be determined by a single number or even by a pair of numbers, as in plane geometry. Instead, three numerical measures are required to specify its position completely.
The coordinates of a point in space are defined as an ordered triplet of real numbers \(x,\,y,\,z\), where each number represents the directed distance of the point from one of the three coordinate planes. The number \(x\) denotes the perpendicular distance of the point from the \(yz\)-plane measured parallel to the \(x\)-axis, \(y\) denotes the perpendicular distance from the \(zx\)-plane measured parallel to the \(y\)-axis, and \(z\) denotes the perpendicular distance from the \(xy\)-plane measured parallel to the \(z\)-axis.
Thus, the ordered triplet \(x,\,y,\,z\) uniquely determines the position of a point in three-dimensional space with respect to the chosen coordinate axes and forms the basis for studying geometrical figures and algebraic relations in three-dimensional geometry.
The sign of the coordinates of a point determine the octant in which the point lies. The following table shows the signs of the coordinates in eight octants.
| \(\dfrac{\text{Octant}}{\text{Coordinate}}\) | I | II | III | IV | V | VI | VII | VIII |
|---|---|---|---|---|---|---|---|---|
| \(x\) | \(+\) | \(-\) | \(-\) | \(+\) | \(+\) | \(-\) | \(-\) | \(+\) |
| \(y\) | \(+\) | \(+\) | \(-\) | \(-\) | \(+\) | \(+\) | \(-\) | \(-\) |
| \(z\) | \(+\) | \(+\) | \(+\) | \(+\) | \(-\) | \(-\) | \(-\) | \(-\) |
Distance between Two Points
In three-dimensional geometry, the distance between two points represents the length of the straight line segment joining them in space. Since a point in space is specified by three coordinates, the distance formula must account for variations along all three mutually perpendicular directions. This concept extends the idea of distance in a plane to space in a natural and logical manner.
Definition
The distance between two points A with coordinates \( (x_1, y_1, z_1) \) and B with coordinates \( (x_2, y_2, z_2) \) is defined as the length of the line segment joining A and B. This distance is denoted by \( AB \).
Derivation of the Distance Formula
Consider two points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) in space. From point B, draw perpendiculars to the coordinate planes so that a rectangular box is formed with edges parallel to the coordinate axes. Let the projections of B on the \( xy \)-plane be used to visualize the horizontal and vertical separations between the points.
The difference in the x-coordinates gives the distance measured parallel to the x-axis, the difference in the y-coordinates gives the distance measured parallel to the y-axis, and the difference in the z-coordinates gives the distance measured parallel to the z-axis.
Thus, the lengths of the edges of the rectangular box are \( |x_2 - x_1| \), \( |y_2 - y_1| \), and \( |z_2 - z_1| \).
Applying the theorem of Pythagoras successively, first in the horizontal plane and then in space, we obtain the required distance.
Proof
Let the distance between the projections of points A and B on the \( xy \)-plane be denoted by \( d \). Using the distance formula in a plane,
\[ \begin{aligned} d^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \end{aligned} \]Now, considering the vertical separation along the z-axis, the distance AB is the hypotenuse of a right-angled triangle whose other sides are \( d \) and \( |z_2 - z_1| \).
\[ \begin{aligned} AB^2 &= d^2 + (z_2 - z_1)^2 \\ &= (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 \end{aligned} \]Taking the positive square root, the distance between the two points is
\[ \begin{aligned} \boxed{\bbox[blue]{\;\;AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\;\;}} \end{aligned} \]Hence, the distance formula for two points in three-dimensional space is established.
Important Aspects
The distance between two points is always a non-negative real number and becomes zero only when the two points coincide. The formula remains valid irrespective of the orientation of the line joining the points, as it depends solely on coordinate differences. If both points lie in the same coordinate plane or along a coordinate axis, the formula naturally reduces to the corresponding two-dimensional or one-dimensional distance formula.
This concept forms the foundation for studying further ideas such as direction ratios, direction cosines, section formula, and equations of lines and planes in space.
Example-1
In Fig 11.3, if P is (2,4,5), find the coordinates of F.
Solution
From the given figure, the point P has coordinates \( (2, 4, 5) \). This means that the distances of P measured parallel to the \(x\)-axis, \(y\)-axis, and \(z\)-axis are 2 units, 4 units, and 5 units respectively.
The point F lies on the face of the rectangular solid where the distance measured parallel to the \(y\)-axis is zero. Hence, the \(y\)-coordinate of F must be zero, while the distances measured parallel to the \(x\)-axis and the \(z\)-axis remain the same as those of point P.
Therefore, the coordinates of point F are obtained by keeping the \(x\)- and \(z\)-coordinates unchanged and replacing the \(y\)-coordinate by zero
\[ \begin{aligned} F &= (2, 0, 5) \end{aligned} \]Example-2
Find the octant in which the points (–3,1,2) and (–3,1,– 2) lie.
Solution
Consider the point \( (-3, 1, 2) \). Here, the x-coordinate is negative, the y-coordinate is positive, and the z-coordinate is positive. An octant is identified by the signs of the coordinates of the point. Since the signs are \( (-, +, +) \), the point lies in the second octant
Now consider the point \( (-3, 1, -2) \). In this case, the x-coordinate is negative, the y-coordinate is positive, and the z-coordinate is negative. Thus, the signs of the coordinates are \( (-, +, -) \), which corresponds to the sixth octant
\[ \begin{aligned} (-3, 1, 2) &\text{ lies in the second octant} \\ (-3, 1, -2) &\text{ lies in the sixth octant} \end{aligned}Example-3
Find the distance between the points P(1, –3, 4) and Q (– 4, 1, 2).
Solution
The coordinates of the given points are \( P(1, -3, 4) \) and \( Q(-4, 1, 2) \). The distance between two points in space is obtained by finding the length of the line segment joining them using the three-dimensional distance formula
\[ \begin{aligned} P &= (1, -3, 4) \\ Q &= (-4, 1, 2) \\ d &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \\ &= \sqrt{(-4 - 1)^2 + (1 + 3)^2 + (2 - 4)^2} \\ &= \sqrt{5^2 + 4^2 + 2^2} \\ &= \sqrt{25 + 16 + 4} \\ &= \sqrt{45} \\ &= 3\sqrt{5} \end{aligned} \]Hence, the distance between the points P and Q is \( 3\sqrt{5} \) units
Example-4
Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear.
Solution
The coordinates of the given points are \( P(-2, 3, 5) \), \( Q(1, 2, 3) \), and \( R(7, 0, -1) \). To show that these three points are collinear, we verify whether the sum of the lengths of two line segments joining the points is equal to the length of the third segment
\[ \begin{aligned} P &= (-2, 3, 5) \\ Q &= (1, 2, 3) \\ R &= (7, 0, -1) \end{aligned} \]First, we find the distance between points P and Q using the distance formula
\[ \begin{aligned} PQ &= \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} \\ &= \sqrt{3^2 + (-1)^2 + (-2)^2} \\ &= \sqrt{9 + 1 + 4} \\ &= \sqrt{14} \end{aligned} \]Next, we calculate the distance between points Q and R
\[ \begin{aligned} QR &= \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} \\ &= \sqrt{6^2 + (-2)^2 + (-4)^2} \\ &= \sqrt{36 + 4 + 16} \\ &= \sqrt{56} \\ &= 2\sqrt{14} \end{aligned} \]Finally, we determine the distance between points P and R
\[ \begin{aligned} PR &= \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} \\ &= \sqrt{9^2 + (-3)^2 + (-6)^2} \\ &= \sqrt{81 + 9 + 36} \\ &= \sqrt{126} \\ &= 3\sqrt{14} \end{aligned} \]Since the sum of the distances \( PQ \) and \( QR \) is equal to the distance \( PR \), that is
\[ \begin{aligned} \sqrt{14} + 2\sqrt{14} &= 3\sqrt{14} \end{aligned} \]it follows that the points P, Q, and R lie on the same straight line and are therefore collinear
Example-5
Are the points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the vertices of a right angled triangle?
Solution
Let us assume that the points A, B, and C are the vertices of a right angled triangle. In such a case, the square of the length of the longest side must be equal to the sum of the squares of the lengths of the other two sides, according to the Pythagoras theorem
\[ \begin{aligned} A &= (3, 6, 9) \\ B &= (10, 20, 30) \\ C &= (25, -41, 5) \end{aligned} \]First, we find the square of the distance between points A and B
\[ \begin{aligned} AB^2 &= (10 - 3)^2 + (20 - 6)^2 + (30 - 9)^2 \\ &= 7^2 + 14^2 + 21^2 \\ &= 49 + 196 + 441 \\ &= 686 \end{aligned} \]Next, we find the square of the distance between points B and C
\[ \begin{aligned} BC^2 &= (25 - 10)^2 + (-41 - 20)^2 + (5 - 30)^2 \\ &= 15^2 + (-61)^2 + (-25)^2 \\ &= 225 + 3721 + 625 \\ &= 4571 \end{aligned} \]Now, we calculate the square of the distance between points C and A
\[ \begin{aligned} CA^2 &= (25 - 3)^2 + (-41 - 6)^2 + (5 - 9)^2 \\ &= 22^2 + (-47)^2 + (-4)^2 \\ &= 484 + 2209 + 16 \\ &= 2709 \end{aligned} \]Here, \( BC^2 \) is the greatest among the three values. For a right angled triangle, it should be equal to the sum of the squares of the other two sides. However,
\[ \begin{aligned} AB^2 + CA^2 &= 686 + 2709 \\ &= 3395 \\ 4571 &\neq 3395 \end{aligned} \]Hence, the points A, B, and C do not form the vertices of a right angled triangle
Example-6
Find the equation of set of points \(P\) such that \(PA^2 + PB^2 = 2k^2\), where \(A\) and \(B\) are the points (3, 4, 5) and (–1, 3, –7), respectively.
Solution
Let the coordinates of the variable point P be \( (x, y, z) \). The given fixed points are \( A(3, 4, 5) \) and \( B(-1, 3, -7) \). Using the distance formula in three dimensional space, we express the squares of the distances \( PA \) and \( PB \) in terms of \( x, y, z \)
\[ \begin{aligned} A &= (3, 4, 5) \\ B &= (-1, 3, -7) \\ P &= (x, y, z) \end{aligned} \]The square of the distance of point P from A is
\[ \begin{aligned} PA^2 &= (x - 3)^2 + (y - 4)^2 + (z - 5)^2 \end{aligned} \]Similarly, the square of the distance of point P from B is
\[ \begin{aligned} PB^2 &= (x + 1)^2 + (y - 3)^2 + (z + 7)^2 \end{aligned} \]According to the given condition, the sum of these squares is equal to \( 2k^2 \)
\[ \begin{aligned} (x - 3)^2 + (y - 4)^2 + (z - 5)^2 + (x + 1)^2 + (y - 3)^2 + (z + 7)^2 &= 2k^2 \end{aligned} \]Grouping like terms and expanding
\[ \begin{aligned} (x - 3)^2 + (x + 1)^2 &= x^2 - 6x + 9 + x^2 + 2x + 1 = 2x^2 - 4x + 10 \\ (y - 4)^2 + (y - 3)^2 &= y^2 - 8y + 16 + y^2 - 6y + 9 = 2y^2 - 14y + 25 \\ (z - 5)^2 + (z + 7)^2 &= z^2 - 10z + 25 + z^2 + 14z + 49 = 2z^2 + 4z + 74 \end{aligned} \]Substituting these results, we obtain
\[ \begin{aligned} 2x^2 - 4x + 2y^2 - 14y + 2z^2 + 4z + 109 &= 2k^2 \\ 2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z &= 2k^2 - 109 \end{aligned} \]This equation represents the required locus of the point P
Example-7
Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
Solution
To show that \(A(1,2,3)\), \(B(-1,-2,-1)\), \(C(2,3,2)\) and \(D(4,7,6)\) are the vertices of a parallelogram but not of a rectangle
$$\begin{aligned} A(1,2,3)\\ B(-1,-2,-1)\\ C(2,3,2)\\ D(4,7,6)\\ AB=\sqrt{(-1-1)^2+(-2-2)^2+(-1-3)^2}\\ =\sqrt{(-2)^2+(-4)^2+(-4)^2}\\ =\sqrt{4+16+16}\\ =\sqrt{36}\\ =6\\ BC=\sqrt{(2-(-1))^2+(3-(-2))^2+(2-(-1))^2}\\ =\sqrt{3^2+5^2+3^2}\\ =\sqrt{9+25+9}\\ =\sqrt{43}\\ CD=\sqrt{(4-2)^2+(7-3)^2+(6-2)^2}\\ =\sqrt{2^2+4^2+4^2}\\ =\sqrt{4+16+16}\\ =\sqrt{36}\\ =6\\ AD=\sqrt{(4-1)^2+(7-2)^2+(6-3)^2}\\ =\sqrt{3^2+5^2+3^2}\\ =\sqrt{9+25+9}\\ =\sqrt{43}\\ AB=CD\\ BC=AD \end{aligned}$$
Since both pairs of opposite sides are equal, the quadrilateral \(ABCD\) is a parallelogram
A parallelogram is a rectangle if and only if its diagonals are equal, so we now compare diagonals \(AC\) and \(BD\)
$$\begin{aligned} AC=\sqrt{(2-1)^2+(3-2)^2+(2-3)^2}\\ =\sqrt{1^2+1^2+(-1)^2}\\ =\sqrt{3}\\ BD=\sqrt{(4-(-1))^2+(7-(-2))^2+(6-(-1))^2}\\ =\sqrt{5^2+9^2+7^2}\\ =\sqrt{25+81+49}\\ =\sqrt{155}\\ AC\ne BD \end{aligned}$$
Since the diagonals are not equal, \(ABCD\) is not a rectangle
Example-8
Find the equation of the set of the points P such that its distances from the points A (3, 4, –5) and B (– 2, 1, 4) are equal.
Solution
Let coordinates of \(P\) be \((x, y, z)\)
$$\begin{aligned} P(x,y,z)\\ A(3,4,-5)\\ B(-2,1,4)\\ PA=\sqrt{(3-x)^2+(4-y)^2+(-5-z)^2}\\ =\sqrt{(3-x)^2+(4-y)^2+(5+z)^2}\\ PB=\sqrt{(x+2)^2+(y-1)^2+(z-4)^2}\\ PA=PB\\ \sqrt{(3-x)^2+(4-y)^2+(5+z)^2}=\sqrt{(x+2)^2+(y-1)^2+(z-4)^2}\\ (3-x)^2+(4-y)^2+(5+z)^2=(x+2)^2+(y-1)^2+(z-4)^2\\ (3-x)^2-(x+2)^2+(4-y)^2-(y-1)^2+(5+z)^2-(z-4)^2=0\\ 9+x^2-6x-x^2-4x-4=-10x+5\\ 16+y^2-8y-y^2+2y-1=-6y+15\\ 25+z^2+10z-z^2+8z-16=18z+9\\ \Rightarrow -10x-6y+18z+29=0\\ \Rightarrow 10x+6y-18z-29=0 \end{aligned}$$
Hence, the required equation of the locus of point \(P\), which is equidistant from points \(A\) and \(B\), is \(10x+6y-18z-29=0\)
Example-9
The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, –5, 7) and (–1, 7, – 6), respectively, find the coordinates of the point C.
Solution
Centroid of triangle is \((1,1,1)\) A \((3,-5,7)\) B \((-1,7,-6)\) Let \(C(x,y,z)\)
$$\begin{aligned} \frac{3+(-1)+x}{3}=1\\ \frac{2+x}{3}=1\\ 2+x=3\\ x=1\\ \frac{-5+7+y}{3}=1\\ \frac{2+y}{3}=1\\ 2+y=3\\ y=1\\ \frac{7+(-6)+z}{3}=1\\ \frac{1+z}{3}=1\\ 1+z=3\\ z=2\\ (x,y,z)=(1,1,2) \end{aligned}$$
Hence, the coordinates of vertex \(C\) are \((1,1,2)\)