1. Coordinate axes and planes in 3D

Chapter 12

Introduction to Three-Dimensional Geometry

Coordinate axes and planes in 3D:

Let Consider three planes intersecting at a point O such that these three planes are mutually perpendicular to each other.

These three planes intersect along the lines XOX, YOY and ZOZ, called the x, y and z-axes, respectively.

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,  XOYZ, XOYZ,  XOYZ, XOYZ,  XOYZ,  XOYZ and XOYZ. and denoted by I, II, III,.. ..., VIII , respectively.

Coordinates of a Point in Space

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,respectively.

In three-dimensional geometry, the coordinates of a point P is always written in the form of P(x, y, z), where x, y and z are the distances of the point, from the YZ, ZX and XY-planes.

  • The coordinates of any point at the origin is (0,0,0)
  • The coordinates of any point on the x-axis is in the form of (x,0,0)
  • The coordinates of any point on the y-axis is in the form of (0,y,0)
  • The coordinates of any point on the z-axis is in the form of (0,0,z)
  • The coordinates of any point on the XY-plane is in the form (x, y, 0)
  • The coordinates of any point on the YZ-plane is in the form (0, y, z)
  • The coordinates of any point on the ZX-plane is in the form (x, 0, z)

Sign of Coordinates in Different Octants:

The sign (+ or -) of the coordinates of a point determines the octant in which the point lies.

Example: In Fig 12.3, 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  Find the octant in which the points (–3,1,2) and (–3,1,– 2) lie.

Solution From the Table 12.1, the point (–3,1, 2) lies in second octant and the point

(–3, 1, – 2) lies in octant VI.

2. Distance between 2 points, section formula

Distance between 2 points, section formula

Distance between Two Points

If P (x1, y1, z1) and Q (x2, y2, z2) are the two points, then the distance between P and Q is given by:

Now PA = y2 – y1, AN = x2 – x1 and NQ = z2 – z1

Hence PQ2 = (x2 – x1)2 + (y2 – y1) 2+ (z2 – z1) 2

Therefore PQ =  [(x2 – x1)2 + (y2 – y1) 2+ (z2 – z1) 2 ]This gives us the distance between two points (x1, y1, z1) and (x2, y2, z2).

In particular, if x1 = y1 = z1 = 0, i.e., point P is origin O, then OQ =  √(x22 + y22 + z22)

which gives the distance between the origin O and any point Q (x2, y2, z2).

PQ=(x2-x1)2+(y2-y1)2+(z2-z1)2

Section Formula

Example : Find the equation of set of points P such that PA2 + PB2 = 2k2, 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 PA2 = (x – 3) 2 + (y – 4) 2 + ( z – 5) 2

PB2 = (x + 1) 2 + (y – 3) 2 + (z + 7) 2

By the given condition PA2 + PB2 = 2k2, we have

(x – 3) 2 + (y – 4) 2 + (z – 5) 2 + (x + 1) 2 + (y – 3) 2 + (z + 7) 2 = 2k2

i.e., 2x2 + 2y2 + 2z2 – 4x – 14y + 4z = 2k2 – 109.

 Let the two given points be P(x1, y1, z1) and Q (x2, y2, z2). Let the point R (x, y, z) divide PQ in the given ratio m : n internally

To determine the coordinates of the point P, the following steps are followed:

  • Draw AL, PN, and BM perpendicular to XY plane such that AL || PN || BM as shown above.
  • The points L, M and N lie on the straight line formed due to the intersection of a plane containing AL, PN and BM and XY- plane.
  • From point P, a line segment ST is drawn such that it is parallel to LM.
  • ST intersects AL externally at S, and it intersects BM at T internally.

Since ST is parallel to LM and AL || PN || BM, therefore, the quadrilaterals LNPS and NMTP qualify as parallelograms.

Also, ∆ASP ~∆BTP therefore,

Rearranging the above equation we get,

Sectional Formula (Internally)

Thus, the coordinates of the point P(x, y, z) dividing the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m:n internally are given by:

Sectional Formula (Externally)

If the given point P divides the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) externally in the ratio m:n, then the coordinates of P are given by replacing n with –n as:

This represents the section formula for three dimension geometry.

If the point P divides the line segment joining points A and B internally in the ratio k:1, then the coordinates of point P will be

What if the point P dividing the line segment joining points A(x1, y1, z1) and B(x2, y2, z2) is the midpoint of line segment AB?

In that case, if P is the midpoint, then P divides the line segment AB in the ratio 1:1, i.e. m=n=1. Coordinates of point P will be given as:

Therefore, the coordinates of the midpoint of line segment joining points A(x1, y1, z1) and B(x2, y2, z2) are given by,

If the point R divides PQ externally in the ratio m : n, then its coordinates are obtained by replacing n by – n so that coordinates of point R will be

The above procedure can be repeated by drawing perpendiculars to XZ and YZ- planes to get the x and y coordinates of the point P that divides the line segment AB in the ratio m:n internally.

Sectional Formula (Internally)

Thus, the coordinates of the point P(x, y, z) dividing the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m:n internally are given by:

Sectional Formula (Externally)

If the given point P divides the line segment joining the points A(x1, y1, z1) and B(x2, y2, z2) externally in the ratio m:n, then the coordinates of P are given by replacing n with –n as:

This represents the section formula for three dimension geometry.

If the point P divides the line segment joining points A and B internally in the ratio k:1, then the coordinates of point P will be

What if the point P dividing the line segment joining points A(x1, y1, z1) and B(x2, y2, z2) is the midpoint of line segment AB?

In that case, if P is the midpoint, then P divides the line segment AB in the ratio 1:1, i.e. m=n=1.

Coordinates of point P will be given as:

Therefore, the coordinates of the midpoint of line segment joining points A(x1, y1, z1) and B(x2, y2, z2) are given by,

If the point R divides PQ externally in the ratio m : n, then its coordinates are obtained by replacing n by – n so that coordinates of point R will be

3. Coordinates of a points

Coordinates of a points

Having chosen a fixed 3D coordinate system in the space, we can associate a given point in the space using three coordinates (x,y,z) and conversely, given a triplet of three numbers (x, y, z), how, we locate a point in the space.

Locate the point “x” on the X-axis. From the point x, moving parallel to the Y-axis, locate the point “y”. Similarly, from the determined point, moving parallel to the Z-axis, locate the point “z”. This is the final coordinate point in the three-dimensional plane, which we are looking for.

when the co-ordinates of the point are given, then we have to draw three planes parallel to XY, YZ and ZX plane meeting the three axes in points A, B and C as shown in the figure. Let OA = x , OB = y and OC = z. Then the coordinates of the point are given as (x,y,z).

  The above the co-ordinates of P are given by (x, y, z). The coordinates of the origin O is (0, 0, 0) Also the coordinates of the point A is given by (x, 0, 0)as A lies completely on the x-axis. Similarly, the coordinates of any point on y-axis is given as (0, y, 0) and on the z-axis, the coordinates are given as (0, 0, z). Also the coordinates of any point in three planes XY, YZ and ZX will be (x,y,0), (0,y,z) and (x,0,z) respectively.

Example: The coordinates of a point are (3, -2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.

Solution:

Given:

Point (3, -2, 5)

The Absolute value of any point(x, y, z) is given by,

√(x2 + y2 + z2)

We need to make sure that absolute value to be the same for all points.

So let the point A(3, -2, 5)

Remaining 7 points are:

Point B(3, 2, 5) (By changing the sign of y coordinate)

Point C(-3, -2, 5) (By changing the sign of x coordinate)

Point D(3, -2, -5) (By changing the sign of z coordinate)

Point E(-3, 2, 5) (By changing the sign of x and y coordinate)

Point F(3, 2, -5) (By changing the sign of y and z coordinate)

Point G(-3, -2, -5) (By changing the sign of x and z coordinate)

Point H(-3, 2, -5) (By changing the sign of x, y and z coordinate)

Example: Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Solution:

Given:

The points (2, 4, 5) and (3, 5, 4)

By using the section formula,

We know X coordinate is always 0 on yz-plane

So, let Point C(0, y, z), and let C divide AB in ratio k: 1

Then, m = k and n = 1

A(2, 4, 5) and B(3, 5, 4)

The coordinates of C are:

On comparing we get,

[3k + 2] / [k + 1] = 0

3k + 2 = 0(k + 1)

3k + 2 = 0

3k = – 2

k = -2/3

We can say that, C divides AB externally in ratio 2: 3

Question 1:

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Solution:

Assume that P (x, y, z) be the point that is equidistant from two points A(1, 2, 3) and B(3, 2, –1).

Thus, we can say that, PA = PB

Take square on both the sides, we get

PA= PB2

It means that,

(x-1)2 + (y-2)2+(z-3)2 = (x-3)2+(y-2)2+(z+1)2

x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1

Now, simplify the above equation, we get:

–2x –4y – 6z + 14 = –6x – 4y + 2z + 14

– 2x – 6z + 6x – 2z = 0

4x – 8z = 0

x – 2z = 0

Hence, the required equation for the set of points is x – 2z = 0.

Question 2:

Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q

divides PR.

Solution:

Assume that the point Q (5, 4, –6) divide the line segment joining points P (3, 2, –4) and R (9, 8, –10) in the ratio k:1.

Therefore, by using the section formula, we can write it as:

(5, 4, -6) = [ (k(9)+3)/(k+1), (k(8)+2)/(k+1), (k(-10)-4)/(k+1)]

(9k+3)/(k+1) = 5

Now, bring the L.H.S denominator to the R.H.S and multiply it

9k+3 = 5k+5

Now, simplify the equation to find the value of k.

4k= 2

k = 2/4

k=½

Therefore, the value of k is ½.

Hence, the point Q divides PR in the ratio of 1:2

Question 3:

Prove that the points: (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right-angled triangle

Solution:

Let the given points be A = (0, 7, 10), B = (–1, 6, 6), and C = (–4, 9, 6)

Now, find the distance between the points

Finding for AB:

AB = √ [(-1-0)2 + (6-7)+(6-10)2]

AB = √ [(-1)2 + (-1)+(-4)2]

AB = √(1+1+16)

AB = √18

AB = 3√2 …. (1)

Finding for BC:

BC= √ [(-4+1)2 + (9-6)+(6-6)2]

BC = √ [(-3)2 + (3)+(-0)2]

BC = √(9+9)

BC = √18

BC = 3√2 …..(2)

Finding for CA:

CA= √ [(0+4)2 + (7-9)+(10-6)2]

CA = √ [(4)2 + (-2)+(4)2]

CA = √(16+4+16)

CA = √36

CA = 6 …..(3)

Now, by Pythagoras theorem,

AC2 = AB2 + BC…..(4)

Now, substitute (1),(2), and (3) in (4), we get:

6= ( 3√2)2 + ( 3√2)2

36 = 18+18

36 = 36

The given points obey the condition of Pythagoras Theorem.

Hence, the given points are the vertices of a right-angled triangle.

Question 4:

Calculate the perpendicular distance of the point P(6, 7, 8) from the XY – Plane.

(a)8  (b)7   (c)6  (d) None of the above

Solution:

A correct answer is an option (A)

Explanation:

Assume that A be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane and the distance of this foot A from P is z-coordinate of P, i.e., 8 units

Hence, the correct answer is an option (a)

Question 5:

If a parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes, then find the length of edges of a parallelopiped and length of the diagonal

Solution:

Let A = (2, 3, 5), B = (5, 9, 7)

To find the length of the edges of a parallelopiped = 5 – 2, 9 – 3, 7 – 5

It means that 3, 6, 2.

Now, to find the length of a diagonal = √(32 + 62 + 22)

= √(9+36+4)

= √49

= 7

Therefore, the length of a diagonal of a parallelopiped is 7 units.