Chapter 11 Introduction To Three Dimensional Geometry
Mathematics is both the queen and the hand-maiden of
all sciences - E.T. BELL
11.1 Introduction
You may recall that to locate the position of a point in a plane, we need two intersecting mutually perpendicular lines in the plane. These lines are called the coordinate axes and the two numbers are called the coordinates of the point with respect to the axes. In actual life, we do not have to deal with points lying in a plane only. For example, consider the position of a ball thrown in space at different points of time or the position of an aeroplane as it flies from one place to another at different times during its flight.
Similarly, if we were to locate the position of the lowest tip of an electric bulb hanging from the ceiling of a room or the position of the central tip of the ceiling fan in a room, we will not only require the perpendicular distances of the point to be located from two perpendicular walls of the room but also the height of the point from the floor of the room. Therefore, we need not only two but three numbers representing the perpendicular distances of the point from three mutually perpendicular planes, namely the floor of the room and two adjacent walls of the room. The three numbers representing the three distances are called the coordinates of the point with reference to the three coordinate planes. So, a point in space has three coordinates. In this Chapter, we shall study the basic concepts of geometry in three dimensional space.*
11.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space
Consider three planes intersecting at a point $O$ such that these three planes are mutually perpendicular to each other (Fig 11.1). These three planes intersect along the lines $X^{\prime} OX, Y^{\prime} OY$ and $Z^{\prime} OZ$, called the $x, y$ and $z$-axes, respectively. We may note that these lines are mutually perpendicular to each other. These lines constitute the rectangular coordinate system. The planes XOY, YOZ and ZOX, called, respectively the XY-plane, YZ-plane and the ZX-plane, are known as the three coordinate planes. We take the XOY plane as the plane of the paper and the
Fig 11.1 line $Z^{\prime} OZ$ as perpendicular to the plane $XOY$. If the plane of the paper is considered as horizontal, then the line $Z^{\prime} OZ$ will be vertical. The distances measured from XY-plane upwards in the direction of $OZ$ are taken as positive and those measured downwards in the direction of $OZ^{\prime}$ are taken as negative. Similarly, the distance measured to the right of $ZX$-plane along $OY$ are taken as positive, to the left of ZX-plane and along $O Y^{\prime}$ as negative, in front of the YZ-plane along $O X$ as positive and to the back of it along $OX^{\prime}$ as negative. The point $O$ is called the origin of the coordinate system. The three coordinate planes divide the space into eight parts known as octants. These octants could be named as $XOYZ, X^{\prime} OYZ, X^{\prime} OY Y, XOY ’ Z, XOYZ$, $X^{\prime} OYZ, X^{\prime} OY^{\prime} Z^{\prime}$ and $XOY^{\prime} Z^{\prime}$. and denoted by I, II, III, …, VIII, respectively.
11.3 Coordinates of a Point in Space
Having chosen a fixed coordinate system in the space, consisting of coordinate axes, coordinate planes and the origin, we now explain, as to how, given a point in the space, we associate with it three coordinates $(x, y, z)$ and conversely, given a triplet of three numbers $(x, y, z)$, how, we locate a point in the space.
Fig 11.2
Given a point $P$ in space, we drop a $\mathbf{X}$ perpendicular PM on the XY-plane with M as th foot of this perpendicular (Fig 11.2). Then, from the point M, we draw a perpendicular ML to the $x$-axis, meeting it at L. Let OL be $x, LM$ be $y$ and MP be $z$. Then $x, y$ and $z$ are called the $x, y$ and $z$ coordinates, respectively, of the point $P$ in the space. In Fig 11.2, we may note that the point $P(x, y, z)$ lies in the octant XOYZ and so all $x, y$, $z$ are positive. If $P$ was in any other octant, the signs of $x, y$ and $z$ would change accordingly. Thus, to each point $P$ in the space there corresponds an ordered triplet $(x, y, z)$ of real numbers.
Conversely, given any triplet $(x, y, z)$, we would first fix the point $L$ on the $x$-axis corresponding to $x$, then locate the point $M$ in the XY-plane such that $(x, y)$ are the coordinates of the point M in the XY-plane. Note that LM is perpendicular to the $x$-axis or is parallel to the $y$-axis. Having reached the point M, we draw a perpendicular MP to the XY-plane and locate on it the point $P$ corresponding to $z$. The point $P$ so obtained has then the coordinates $(x, y, z)$. Thus, there is a one to one correspondence between the points in space and ordered triplet $(x, y, z)$ of real numbers.
Alternatively, through the point $P$ in the space, we draw three planes parallel to the coordinate planes, meeting the $x$-axis, $y$-axis and $z$-axis in the points $A, B$ and $C$, respectively (Fig 11.3). Let $OA=x, OB=y$ and $OC=z$. Then, the point $P$ will have the coordinates $x, y$ and $z$ and we write $P(x, y, z)$. Conversely, given $x, y$ and $z$, we locate the three points $A, B$ and $C$ on the three coordinate axes. Through the points $A, B$ and $C$ we draw planes parallel to the YZ-plane, ZX-plane and XY-plane,
Fig 11.3
respectively. The point of interesection of these three planes, namely, ADPF, BDPE and CEPF is obviously the point $P$, corresponding to the ordered triplet $(x, y, z)$. We observe that if $P(x, y, z)$ is any point in the space, then $x, y$ and $z$ are perpendicular distances from YZ, ZX and XY planes, respectively.
Note - The coordinates of the origin $O$ are $(0,0,0)$. The coordinates of any point on the $x$-axis will be as $(x, 0,0)$ and the coordinates of any point in the YZ-plane will be as $(0, y, z)$.
Remark 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.
Table 11.1
| Octants/des | I | II | III | IV | V | VI | VII | VIII |
|---|---|---|---|---|---|---|---|---|
| cootdinate | + | - | - | + | + | - | - | + |
| $x$ | + | + | - | - | + | + | - | - |
| $y$ | + | + | + | + | - | - | - | - |
Example 1 In Fig 11.3, if $P$ is $(2,4,5)$, find the coordinates of $F$.
Solution For the point $F$, the distance measured along OY is zero. Therefore, the coordinates of $F$ are $(2,0,5)$.
Example 2 Find the octant in which the points $(-3,1,2)$ and $(-3,1,-2)$ lie.
Solution From the Table 11.1, the point $(-3,1,2)$ lies in second octant and the point $(-3,1,-2)$ lies in octant VI.
EXERCISE 11.1
1. A point is on the $x$-axis. What are its $y$-coordinate and $z$-coordinates?
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Answer :
If a point is on the $x$-axis, then its $y$-coordinates and $z$-coordinates are zero.
2. A point is in the XZ-plane. What can you say about its $y$-coordinate?
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Answer :
If a point is in the XZ plane, then its $y$-coordinate is zero.
3. Name the octants in which the following points lie:
$ \begin{aligned} & (1,2,3),(4,-2,3),(4,-2,-5),(4,2,-5),(-4,2,-5),(-4,2,5), \\ & (-3,-1,6)(-2,-4,-7) . \end{aligned} $
Show Answer
Answer :
The $x$-coordinate, $y$-coordinate, and $z$-coordinate of point $(1,2,3)$ are all positive. Therefore, this point lies in octant $\mathbf{I}$.
The $x$-coordinate, $y$-coordinate, and $z$-coordinate of point $(4,-2,3)$ are positive, negative, and positive respectively. Therefore, this point lies in octant IV.
The $x$-coordinate, $y$-coordinate, and $z$-coordinate of point $(4,-2,-5)$ are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
The $x$-coordinate, $y$-coordinate, and $z$-coordinate of point $(4,2,-5)$ are positive, positive, and negative respectively. Therefore, this point lies in octant $\mathbf{V}$.
The $x$-coordinate, $y$-coordinate, and $z$-coordinate of point $(-4,2,-5)$ are negative, positive, and negative respectively. Therefore, this point lies in octant VI.
The $x$-coordinate, $y$-coordinate, and $z$-coordinate of point $(-4,2,5)$ are negative, positive, and positive respectively. Therefore, this point lies in octant II.
The $x$-coordinate, $y$-coordinate, and $z$-coordinate of point $(-3,-1,6)$ are negative, negative, and positive respectively. Therefore, this point lies in octant III.
The $x$-coordinate, $y$-coordinate, and $z$-coordinate of point $(-2,-4,-7)$ are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
4. Fill in the blanks:
(i) The $x$-axis and $y$-axis taken together determine a plane known as ____________
(ii) The coordinates of points in the XY-plane are of the form ___________
(iii) Coordinate planes divide the space into ___________ octants.
Show Answer
Answer :
(i) $XY-$ plane
(ii) ${(x, y, 0)}$.
(iii) ${\text{ eight }}$
11.4 Distance between Two Points
We have studied about the distance between two points in two-dimensional coordinate system. Let us now extend this study to three-dimensional system.
Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be two points referred to a system of rectangular axes $OX, OY$ and $OZ$. Through the points $P$ and $Q$ draw planes parallel to the coordinate planes so as to form a rectangular parallelopiped with one diagonal PQ (Fig 11.4).
Fig 11.4
Now, since $\angle PAQ$ is a right $\quad \mathbf{X}$ angle, it follows that, in triangle PAQ,
$ PQ^{2}=PA^{2}+AQ^{2} \quad\quad\quad\quad\quad\quad\quad \ldots(1) $
Also, triangle ANQ is right angle triangle with $\angle ANQ$ a right angle.
Therefore $\quad\quad\quad AQ^{2}=AN^{2}+NQ^{2} \quad\quad\quad\quad\quad\quad\quad \ldots(2)$
From (1) and (2), we have
$$ \mathrm{PQ}^{2}=\mathrm{PA}^{2}+\mathrm{AN}^{2}+\mathrm{NQ}^{2} $$
Now $\quad\quad\quad\mathrm{PA}=y _{2}-y _{1}, \mathrm{AN}=x _{2}-x _{1}$ और $\mathrm{NQ}=z _{2}-z _{1}$
Hence $\quad\quad\quad\mathrm{PQ}^{2}=\left(x _{2}-x _{1}\right)^{2}+\left(y _{2}-y _{1}\right)^{2}+\left(z _{2}-z _{1}\right)^{2}$
Therefore $\quad\quad\quad PQ=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}+(z_2-z_1)^{2}}$
This gives us the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$.
In particular, if $x_1=y_1=z_1=0$, i.e., point $P$ is origin $O$, then $OQ=\sqrt{x_2{ }^{2}+y_2{ }^{2}+z_2{ }^{2}}$, which gives the distance between the origin $O$ and any point $Q(x_2, y_2, z_2)$.
Example 3 Find the distance between the points $P(1,-3,4)$ and $Q(-4,1,2)$.
Solution The distance PQ between the points $P(1,-3,4)$ and $Q(-4,1,2)$ is
$ \begin{aligned} PQ & =\sqrt{(-4-1)^{2}+(1+3)^{2}+(2-4)^{2}} \\ & =\sqrt{25+16+4} \\ & =\sqrt{45}=3 \sqrt{5} \text{ units } \end{aligned} $
Example 4 Show that the points $P(-2,3,5)$, $Q(1,2,3)$ and $R(7,0,-1)$ are collinear.
Solution We know that points are said to be collinear if they lie on a line.
Now,
$ \begin{aligned} & P Q=\sqrt{(1+2)^{2}+(2-3)^{2}+(3-5)^{2}}=\sqrt{9+1+4}=\sqrt{14} \\ & Q R=\sqrt{(7-1)^{2}+(0-2)^{2}+(-1-3)^{2}}=\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14} \end{aligned} $
and
$ P R=\sqrt{(7+2)^{2}+(0-3)^{2}+(-1-5)^{2}}=\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14} $
Thus, $PQ+QR=PR$. Hence, $P, Q$ and $R$ are 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 By the distance formula, we have
$ \begin{aligned} AB^{2} & =(10-3)^{2}+(20-6)^{2}+(30-9)^{2} \\ & =49+196+441=686 \\ BC^{2} & =(25-10)^{2}+(-41-20)^{2}+(5-30)^{2} \\ & =225+3721+625=4571 \end{aligned} $
$ \begin{aligned} CA^{2} & =(3-25)^{2}+(6+41)^{2}+(9-5)^{2} \\ & =484+2209+16=2709 \end{aligned} $
We find that $\quad \quad\quad CA^{2}+AB^{2} \neq BC^{2}$.
Hence, the triangle $ABC$ is not a right angled triangle.
Example 6 Find the equation of set of points $P$ such that $PA^{2}+PB^{2}=2 k^{2}$, where $A$ and $B$ are the points $(3,4,5)$ and $(-1,3,-7)$, respectively.
Solution Let the coordinates of point $P$ be $(x, y, z)$.
Here
$ \begin{aligned} & PA^{2}=(x-3)^{2}+(y-4)^{2}+(z-5)^{2} \\ & PB^{2}=(x+1)^{2}+(y-3)^{2}+(z+7)^{2} \end{aligned} $
By the given condition $PA^{2}+PB^{2}=2 k^{2}$, we have
$ (x-3)^{2}+(y-4)^{2}+(z-5)^{2}+(x+1)^{2}+(y-3)^{2}+(z+7)^{2}=2 k^{2} \\ \text{ i.e., } \quad 2 x^{2}+2 y^{2}+2 z^{2}-4 x-14 y+4 z=2 k^{2}-109 . $
EXERCISE 11.2
1. Find the distance between the following pairs of points:
(i) $(2,3,5)$ and $(4,3,1)$
(ii) $(-3,7,2)$ and $(2,4,-1)$
(iii) $(-1,3,-4)$ and $(1,-3,4)$
(iv) $(2,-1,3)$ and $(-2,1,3)$.
Show Answer
Answer :
The distance between points $P(x_1, y_1, z_1)$ and $P(x_2, y_2, z_2)$ is given
(i) Distance between points $(2,3,5)$ and $(4,3,1)$
$=\sqrt{(4-2)^{2}+(3-3)^{2}+(1-5)^{2}}$
$=\sqrt{(2)^{2}+(0)^{2}+(-4)^{2}}$
$=\sqrt{4+16}$
$=\sqrt{20}$
$=2 \sqrt{5}$
(ii) Distance between points (-3, 7, 2) and (2, 4, -1) $=\sqrt{(2+3)^{2}+(4-7)^{2}+(-1-2)^{2}}$
$=\sqrt{(5)^{2}+(-3)^{2}+(-3)^{2}}$
$=\sqrt{25+9+9}$
$=\sqrt{43}$
(iii) Distance between points (-1, 3, -4) and (1, -3, 4)
$=\sqrt{(1+1)^{2}+(-3-3)^{2}+(4+4)^{2}}$
$=\sqrt{(2)^{2}+(-6)^{3}+(8)^{2}}$
$=\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}$
(iv) Distance between points (2, -1,3) and (-2, 1, 3)
$=\sqrt{(-2-2)^{2}+(1+1)^{2}+(3-3)^{2}}$
$=\sqrt{(-4)^{2}+(2)^{2}+(0)^{2}}$
$=\sqrt{16+4}$
$=\sqrt{20}$
$=2 \sqrt{5}$
2. Show that the points $(-2,3,5),(1,2,3)$ and $(7,0,-1)$ are collinear.
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Answer :
Let points (-2, 3, 5), (1, 2, 3), and (7, 0, -1) be denoted by P, Q, and R respectively.
Points $P, Q$, and $R$ are collinear if they lie on a line.
$ \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} \\ 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} $
$ \begin{aligned} P R= & \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} $
Here, $PQ+QR=\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}=PR$
Hence, points $P(- 2,3,5), Q(1,2,3)$, and $R(7,0$, - 1 $)$ are collinear.
3. Verify the following:
(i) $(0,7,-10),(1,6,-6)$ and $(4,9,-6)$ are the vertices of an isosceles triangle.
(ii) $(0,7,10),(-1,6,6)$ and $(-4,9,6)$ are the vertices of a right angled triangle.
(iii) $(-1,2,1),(1,-2,5),(4,-7,8)$ and $(2,-3,4)$ are the vertices of a parallelogram.
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Answer :
(i) Let points $(0,7$, -10), (1, 6, - 6 ), and (4, 9, -6) be denoted by A, B, and C respectively.
$ \begin{aligned} AB & =\sqrt{(1-0)^{2}+(6-7)^{2}+(-6+10)^{2}} \\ & =\sqrt{(1)^{2}+(-1)^{2}+(4)^{2}} \\ & =\sqrt{1+1+16} \\ & =\sqrt{18} \\ & =3 \sqrt{2} \\ BC & =\sqrt{(4-1)^{2}+(9-6)^{2}+(-6+6)^{2}} \\ & =\sqrt{(3)^{2}+(3)^{2}} \\ & =\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \\ CA & =\sqrt{(0-4)^{2}+(7-9)^{2}+(-10+6)^{2}} \\ & =\sqrt{(-4)^{2}+(-2)^{2}+(-4)^{2}} \\ & =\sqrt{16+4+16}=\sqrt{36}=6 \end{aligned} $
Here, $A B=B C \neq C A$
Thus, the given points are the vertices of an isosceles triangle.
(ii) Let $(0,7,10)$, (-1, 6, 6), and (- $ 4,9,6 )$ be denoted by A, B, and C respectively.
$ \begin{aligned} AB & =\sqrt{(-1-0)^{2}+(6-7)^{2}+(6-10)^{2}} \\ & =\sqrt{(-1)^{2}+(-1)^{2}+(-4)^{2}} \\ & =\sqrt{1+1+16}=\sqrt{18} \\ & =3 \sqrt{2} \\ BC & =\sqrt{(-4+1)^{2}+(9-6)^{2}+(6-6)^{2}} \\ & =\sqrt{(-3)^{2}+(3)^{2}+(0)^{2}} \\ & =\sqrt{9+9}=\sqrt{18} \\ & =3 \sqrt{2} \end{aligned} $
$ \begin{aligned} CA & =\sqrt{(0+4)^{2}+(7-9)^{2}+(10-6)^{2}} \\ & =\sqrt{(4)^{2}+(-2)^{2}+(4)^{2}} \\ & =\sqrt{16+4+16} \\ & =\sqrt{36} \\ & =6 \end{aligned} $
Now, $AB^{2}+BC^{2}=(3 \sqrt{2})^{2}+(3 \sqrt{2})^{2}=18+18=36=AC^{2}$
Therefore, by Pythagoras theorem, $A B C$ is a right triangle.
Hence, the given points are the vertices of a right-angled triangle.
(iii) Let $(-1,2,1),(1, - 2,5),(4, - 7,8)$, and $(2, - 3,4)$ be denoted by $A, B, C$, and $D$ respectively.
$ \begin{aligned} AB & =\sqrt{(1+1)^{2}+(-2-2)^{2}+(5-1)^{2}} \\ & =\sqrt{4+16+16} \\ & =\sqrt{36} \\ & =6 \end{aligned} $
$ \begin{aligned} BC & =\sqrt{(4-1)^{2}+(-7+2)^{2}+(8-5)^{2}} \\ & =\sqrt{9+25+9}=\sqrt{43} \\ CD & =\sqrt{(2-4)^{2}+(-3+7)^{2}+(4-8)^{2}} \\ & =\sqrt{4+16+16} \\ & =\sqrt{36} \\ & =6 \end{aligned} $
$ \begin{aligned} DA & =\sqrt{(-1-2)^{2}+(2+3)^{2}+(1-4)^{2}} \\ & =\sqrt{9+25+9}=\sqrt{43} \end{aligned} $
Here, $A B=C D=6, B C=A D=\sqrt{43}$
Hence, the opposite sides of quadrilateral $A B C D$, whose vertices are taken in order, are equal.
Therefore, $A B C D$ is a parallelogram.
Hence, the given points are the vertices of a parallelogram.
4. Find the equation of the set of points which are equidistant from the points $(1,2,3)$ and $(3,2,-1)$.
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Answer :
Let $P(x, y, z)$ be the point that is equidistant from points $A(1,2,3)$ and $B(3,2, - 1)$.
Accordingly, PA $=$ PB
$\Rightarrow PA^{2}=PB^{2}$
$\Rightarrow(x-1)^{2}+(y-2)^{2}+(z-3)^{2}=(x-3)^{2}+(y-2)^{2}+(z+1)^{2}$
$ x^2-2x+1 +y^2 - 4y +4 + z^2-6z+9 = x^2 -6x+9 + y^2 - 4y +4 +z^2 +2z+1 $
$\Rightarrow$ - $2 x$ - $4 y - 6 z+14=- 6 x$ - $4 y+2 z+14$
$\Rightarrow$ - $2 x - 6 z+6 x - 2 z=0$
$\Rightarrow 4 x - 8 z=0$
$\Rightarrow x - 2 z=0$
Thus, the required equation is $x - 2 z=0$.
5. Find the equation of the set of points $P$, the sum of whose distances from $A(4,0,0)$ and $B(-4,0,0)$ is equal to 10 .
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Answer :
Let the coordinates of $P$ be $(x, y, z)$.
The coordinates of points $A$ and $B$ are $(4,0,0)$ and $(-4,0,0) $ respectively.
It is given that $PA+PB=10$.
$ \begin{aligned} & \Rightarrow \sqrt{(x-4)^{2}+y^{2}+z^{2}}+\sqrt{(x+4)^{2}+y^{2}+z^{2}}=10 \\ & \Rightarrow \sqrt{(x-4)^{2}+y^{2}+z^{2}}=10-\sqrt{(x+4)^{2}+y^{2}+z^{2}} \end{aligned} $
On squaring both sides, we obtain
$ \begin{aligned} & \Rightarrow(x-4)^{2}+y^{2}+z^{2}=100-20 \sqrt{(x+4)^{2}+y^{2}+z^{2}}+(x+4)^{2}+y^{2}+z^{2} \\ & \Rightarrow x^{2}-8 x+16+y^{2}+z^{2}=100-20 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}+x^{2}+8 x+16+y^{2}+z^{2} \\ & \Rightarrow 20 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}=100+16 x \\ & \Rightarrow 5 \sqrt{x^{2}+8 x+16+y^{2}+z^{2}}=(25+4 x) \end{aligned} $
On squaring both sides again, we obtain
$25(x^{2}+8 x+16+y^{2}+z^{2})=625+16 x^{2}+200 x$
$\Rightarrow 25 x^{2}+200 x+400+25 y^{2}+25 z^{2}=625+16 x^{2}+200 x$
$\Rightarrow 9 x^{2}+25 y^{2}+25 z^{2} - 225=0$
Thus, the required equation is $9 x^{2}+25 y^{2}+25 z^{2} - 225=0$.
Miscellaneous Examples
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 ABCD is a parallelogram we need to show opposite side are equal Note that.
$ \begin{aligned} & AB=\sqrt{(-1-1)^{2}+(-2-2)^{2}+(-1-3)^{2}}=\sqrt{4+16+16}=6 \\ & BC=\sqrt{(2+1)^{2}+(3+2)^{2}+(2+1)^{2}}=\sqrt{9+25+9}=\sqrt{43} \\ & CD=\sqrt{(4-2)^{2}+(7-3)^{2}+(6-2)^{2}}=\sqrt{4+16+16}=6 \\ & DA=\sqrt{(1-4)^{2}+(2-7)^{2}+(3-6)^{2}}=\sqrt{9+25+9}=\sqrt{43} \end{aligned} $
Since $A B=C D$ and $B C=A D, A B C D$ is a parallelogram.
Now, it is required to prove that $ABCD$ is not a rectangle. For this, we show that diagonals $AC$ and $BD$ are unequal. We have
$ \begin{aligned} & AC=\sqrt{(2-1)^{2}+(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1+1}=\sqrt{3} \\ & BD \quad=\sqrt{(4+1)^{2}+(7+2)^{2}+(6+1)^{2}}=\sqrt{25+81+49}=\sqrt{155} . \end{aligned} $
Since $A C \neq B D, A B C D$ is not a rectangle.
Note - We can also show that $ABCD$ is a parallelogram, using the property that diagonals $AC$ and $BD$ bisect each other.
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 If $P(x, y, z)$ be any point such that $PA=PB$.
Now $\sqrt{(x-3)^{2}+(y-4)^{2}+(z+5)^{2}} = \sqrt{(x+2)^{2}+(y-1)^{2}+(z-4)^{2}}$
or $\quad\quad (x-3)^{2}+(y-4)^{2}+(z+5)^{2}=(x+2)^{2}+(y-1)^{2}+(z-4)^{2}$
or $\quad \quad 10 x+6 y-18 z-29=0$.
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 Let the coordinates of $C$ be $(x, y, z)$ and the coordinates of the centroid $G$ be $(1,1,1)$. Then
Therefore $\quad \frac{x+3-1}{3}=1$, or $x=1$ $ \begin{array}{ll} \frac{y-5+7}{3}=1, & \text { or } y=1 \\ \frac{z+7-6}{3}=1, & \text { or } z=2 . \end{array} $
Hence, coordinates of $C$ are $(1,1,2)$.
Miscellaneous Exercise on Chapter 11
1. Three vertices of a parallelogram $ABCD$ are $A(3,-1,2), B(1,2,-4)$ and $C(-1,1,2)$. Find the coordinates of the fourth vertex.
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Answer :
The three vertices of a parallelogram $A B C D$ are given as $ A(3, - $ 1, 2 )$, B(1,2,- 4)$, and C (-1, 1, 2). Let the coordinates of the fourth vertex be $D(x, y, z)$.
We know that the diagonals of a parallelogram bisect each other.
Therefore, in parallelogram $A B C D, A C$ and $B D$ bisect each other.
$\therefore$ Mid-point of $A C=$ Mid-point of $B D$
$\Rightarrow(\dfrac{3-1}{2}, \dfrac{-1+1}{2}, \dfrac{2+2}{2})=(\dfrac{x+1}{2}, \dfrac{y+2}{2}, \dfrac{z-4}{2})$
$\Rightarrow(1,0,2)=(\dfrac{x+1}{2}, \dfrac{y+2}{2}, \dfrac{z-4}{2})$
$\Rightarrow \dfrac{x+1}{2}=1, \dfrac{y+2}{2}=0$, and $\dfrac{z-4}{2}=2$
$\Rightarrow x=1, y=- 2$ , and $z=8$
Thus, the coordinates of the fourth vertex are (1, -2, 8).
2. Find the lengths of the medians of the triangle with vertices $A(0,0,6), B(0,4,0)$ and $(6,0,0)$.
Show Answer
Answer :
Let $A D, B E$, and $C F$ be the medians of the given triangle $A B C$.
Since $A D$ is the median, $D$ is the mid-point of $B C$.
$\therefore$ Coordinates of point $D=(\dfrac{0+6}{2}, \dfrac{4+0}{2}, \dfrac{0+0}{2})=(3,2,0)$
$AD=\sqrt{(0-3)^{2}+(0-2)^{2}+(6-0)^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$
Since $BE$ is the median, $E$ is the mid-point of $AC$.
$\therefore$ Coordinates of point $E=(\dfrac{0+6}{2}, \dfrac{0+0}{2}, \dfrac{6+0}{2})=(3,0,3)$
$BE=\sqrt{(3-0)^{2}+(0-4)^{2}+(3-0)^{2}}=\sqrt{9+16+9}=\sqrt{34}$
Since $CF$ is the median, $F$ is the mid-point of $AB$.
$\therefore$ Coordinates of point $F=(\dfrac{0+0}{2}, \dfrac{0+4}{2}, \dfrac{6+0}{2})=(0,2,3)$
Length of $C F=\sqrt{(6-0)^{2}+(0-2)^{2}+(0-3)^{2}}=\sqrt{36+4+9}=\sqrt{49}=7$
Thus, the lengths of the medians of $\triangle A B C$ are $7, \sqrt{34}$, and 7
3. If the origin is the centroid of the triangle $PQR$ with vertices $P(2 a, 2,6)$, $Q(-4,3 b,-10)$ and $R(8,14,2 c)$, then find the values of $a, b$ and $c$.
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Answer :
It is known that the coordinates of the centroid of the triangle, whose vertices are $(x_1, y_1, z_1),(x_2, y_2, z_2)$ and $(x_3, y_3, z_3)$, are $(\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}, \dfrac{z_1+z_2+z_3}{3})$
Therefore, coordinates of the centroid of
$\Delta PQR=(\dfrac{2 a-4+8}{3}, \dfrac{2+3 b+14}{3}, \dfrac{6-10+2 c}{3})=(\dfrac{2 a+4}{3}, \dfrac{3 b+16}{3}, \dfrac{2 c-4}{3})$
It is given that origin is the centroid of $\triangle P Q R$.
$\therefore(0,0,0)=(\dfrac{2 a+4}{3}, \dfrac{3 b+16}{3}, \dfrac{2 c-4}{3})$
$\Rightarrow \dfrac{2 a+4}{3}=0, \dfrac{3 b+16}{3}=0$ and $\dfrac{2 c-4}{3}=0$
$\Rightarrow a=-2, b=-\dfrac{16}{3}$ and $c=2$
Thus, the respective values of $a, b$, and $c$ are
$ -2,-\dfrac{16}{3}, \text{ and } 2 \text{. } $
4. If $A$ and $B$ be the points $(3,4,5)$ and $(-1,3,-7)$, respectively, find the equation of the set of points $P$ such that $PA^{2}+PB^{2}=k^{2}$, where $k$ is a constant.
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Answer :
The coordinates of points A and B are given as $(3,4,5)$ and $( -1,3,-7 )$ respectively.
Let the coordinates of point $P$ be $(x, y, z)$.
On using distance formula, we obtain
$ \begin{aligned} PA^{2} & =(x-3)^{2}+(y-4)^{2}+(z-5)^{2} \\ & =x^{2}+9-6 x+y^{2}+16-8 y+z^{2}+25-10 z \\ & =x^{2}-6 x+y^{2}-8 y+z^{2}-10 z+50 \\ PB^{2} & =(x+1)^{2}+(y-3)^{2}+(z+7)^{2} \\ & =x^{2}+2 x+y^{2}-6 y+z^{2}+14 z+59 \end{aligned} $
Now, if $PA^{2}+PB^{2}=k^{2}$, then
$ \begin{aligned} & (x^{2}-6 x+y^{2}-8 y+z^{2}-10 z+50)+(x^{2}+2 x+y^{2}-6 y+z^{2}+14 z+59)=k^{2} \\ & \Rightarrow 2 x^{2}+2 y^{2}+2 z^{2}-4 x-14 y+4 z+109=k^{2} \\ & \Rightarrow 2(x^{2}+y^{2}+z^{2}-2 x-7 y+2 z)=k^{2}-109 \\ & \Rightarrow x^{2}+y^{2}+z^{2}-2 x-7 y+2 z=\dfrac{k^{2}-109}{2} \\ & \qquad x^{2}+y^{2}+z^{2}-2 x-7 y+2 z=\dfrac{k^{2}-109}{2} . \end{aligned} $
Summary
In three dimensions, the coordinate axes of a rectangular Cartesian coordinate system are three mutually perpendicular lines. The axes are called the $x$, $y$ and $z$-axes.
The three planes determined by the pair of axes are the coordinate planes, called XY, YZ and ZX-planes.
The three coordinate planes divide the space into eight parts known as octants. The coordinates of a point $P$ in three dimensional geometry is always written in the form of triplet like $(x, y, z)$. Here $x, y$ and $z$ are the distances from the YZ, ZX and XY-planes.
(i) Any point on $x$-axis is of the form $(x, 0,0)$
(ii) Any point on $y$-axis is of the form $(0, y, 0)$
(iii) Any point on $z$-axis is of the form $(0,0, z)$.
Distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by
$$ PQ=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}+(z_2-z_1)^{2}} $$
Historical Note
Rene’ Descartes (1596-1650), the father of analytical geometry, essentially dealt with plane geometry only in 1637 . The same is true of his co-inventor Pierre Fermat (1601-1665) and La Hire (1640-1718). Although suggestions for the three dimensional coordinate geometry can be found in their works but no details. Descartes had the idea of coordinates in three dimensions but did not develop it. J.Bernoulli (1667-1748) in a letter of 1715 to Leibnitz introduced the three coordinate planes which we use today. It was Antoinne Parent (1666-1716), who gave a systematic development of analytical solid geometry for the first time in a paper presented to the French Academy in 1700.
L.Euler (1707-1783) took up systematically the three dimensional coordinate geometry, in Chapter 5 of the appendix to the second volume of his “Introduction to Geometry” in 1748.
It was not until the middle of the nineteenth century that geometry was extended to more than three dimensions, the well-known application of which is in the Space-Time Continuum of Einstein’s Theory of Relativity.
