Wednesday, October 9, 2024

Introduction To There Dimensional Geometry

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Chapter 11.1: Coordinate Axes and Coordinates in Space

  • Coordinate Axes: Three mutually perpendicular lines intersecting at a point called the origin.
  • Coordinates of a Point: The ordered triple (x, y, z) representing the position of a point in space relative to the coordinate axes.

Chapter 11.2: Distance Between Two Points

  • Distance Formula: The distance between two points (x1, y1, z1) and (x2, y2, z2) is given by:
    • d = √((x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2)

Chapter 11.3: Section Formula

  • Section Formula: The coordinates of a point dividing a line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m:n are given by:
    • x = (mx2 + nx1) / (m + n)
    • y = (my2 + ny1) / (m + n)
    • z = (mz2 + nz1) / (m + n)

Chapter 11.4: Direction Ratios and Direction Cosines

  • Direction Ratios: Any three non-zero numbers proportional to the direction cosines of a line.
  • Direction Cosines: The cosines of the angles made by a line with the x, y, and z-axes, denoted by l, m, and n, respectively.
  • Relation Between Direction Ratios and Direction Cosines: l^2 + m^2 + n^2 = 1

Key Concepts:

  • Coordinate axes and coordinates in space
  • Distance between two points
  • Section formula
  • Direction ratios and direction cosines
  • Applications of three-dimensional geometry (e.g., finding distances, equations of lines and planes)

Exercise 11.1

1. A point is on the x-axis. What are its y-coordinate and z-coordinates?

Ans : A point on the x-axis has y-coordinate and z-coordinate equal to 0.

2. A point is in the XZ-plane. What can you say about its y-coordinate?

Ans : A point in the XZ-plane has a y-coordinate of 0.

3. Name the octants in which the following points lie: (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (– 2, – 4, –7).

Ans : 

Octant I (x+, y+, z+):

  • (1, 2, 3)

Octant IV (x+, y-, z+):

  • (4, -2, 3)

Octant VIII (x+, y-, z-):

  • (4, -2, -5)

Octant V (x+, y+, z-):

  • (4, 2, -5)

Octant VI (x-, y+, z-):

  • (-4, 2, -5)

Octant II (x-, y+, z+):

  • (-4, 2, 5)

Octant III (x-, y-, z+):

  • (-3, -1, 6)

Octant VII (x-, y-, z-):

  • (-2, -4, -7)

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. 

Ans : 

(i) XY-plane.

(ii) (x, y, 0).  

(iii)eight octants.

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). 

Ans : 

d = √((x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2)

(i) Distance between (2, 3, 5) and (4, 3, 1):

d = √((4 – 2)^2 + (3 – 3)^2 + (1 – 5)^2)

d = √(2^2 + 0^2 + (-4)^2)

d = √(4 + 0 + 16)

d = √20

d = 2√5

(ii) Distance between (-3, 7, 2) and (2, 4, -1):

d = √((2 – (-3))^2 + (4 – 7)^2 + (-1 – 2)^2)

d = √(5^2 + (-3)^2 + (-3)^2)

d = √(25 + 9 + 9)

d = √43

(iii) Distance between (-1, 3, -4) and (1, -3, 4):

d = √((1 – (-1))^2 + (-3 – 3)^2 + (4 – (-4))^2)

d = √(2^2 + (-6)^2 + 8^2)

d = √(4 + 36 + 64)

d = √104

d = 2√26

(iv) Distance between (2, -1, 3) and (-2, 1, 3):

d = √((-2 – 2)^2 + (1 – (-1))^2 + (3 – 3)^2)

d = √((-4)^2 + 2^2 + 0^2)

d = √(16 + 4 + 0)

d = √20

d = 2√5

2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear

Ans : 

Let the points be A(-2, 3, 5), B(1, 2, 3), and C(7, 0, -1).

Calculating the distances:

AB = √((1 – (-2))^2 + (2 – 3)^2 + (3 – 5)^2) = √(9 + 1 + 4) = √14

BC = √((7 – 1)^2 + (0 – 2)^2 + (-1 – 3)^2) = √(36 + 4 + 16) = √56 = 2√14

AC = √((7 – (-2))^2 + (0 – 3)^2 + (-1 – 5)^2) = √(81 + 9 + 36) = √126 = 3√14

Checking the collinearity condition:

AB + BC = AC

√14 + 2√14 = 3√14

Therefore, the points A, B, and C 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

Ans : 

Let’s denote the points as follows:

  • A = (0, 7, -10)
  • B = (1, 6, -6)
  • C = (4, 9, -6)
  • D = (0, 7, 10)
  • E = (-1, 6, 6)
  • F = (-4, 9, 6)
  • P = (-1, 2, 1)
  • Q = (1, -2, 5)
  • R = (4, -7, 8)
  • S = (2, -3, 4)

Recall the conditions for different types of triangles:

  • Isosceles triangle: Two sides are equal.
  • Right-angled triangle: One angle is 90 degrees.
  • Parallelogram: Opposite sides are equal and parallel.

(i) Isosceles Triangle ABC

  • Calculate distances:
    • AB = √((1-0)^2 + (6-7)^2 + (-6+10)^2) = √17
    • BC = √((4-1)^2 + (9-6)^2 + (-6+6)^2) = √18
    • AC = √((4-0)^2 + (9-7)^2 + (-6+10)^2) = √32
  • Conclusion: Since AB = BC, triangle ABC is isosceles.

(ii) Right-angled Triangle DEF

  • Calculate distances:
    • DE = √((-1-0)^2 + (6-7)^2 + (6-10)^2) = √17
    • EF = √((-4+1)^2 + (9-6)^2 + (6-6)^2) = √18
    • DF = √((-4-0)^2 + (9-7)^2 + (6-10)^2) = √32
  • Check Pythagoras Theorem:
    • DE^2 + EF^2 = DF^2
    • 17 + 18 = 32
    • 35 = 32
  • Conclusion: Since the Pythagorean Theorem holds, triangle DEF is right-angled.

(iii) Parallelogram PQRS

  • Calculate distances:
    • PQ = √((1-(-1))^2 + (-2-2)^2 + (5-1)^2) = √32
    • QR = √((4-1)^2 + (-7+2)^2 + (8-5)^2) = √32
    • RS = √((2-4)^2 + (-3+7)^2 + (4-8)^2) = √32
    • SP = √((-1-2)^2 + (2-(-3))^2 + (1-4)^2) = √32
  • Check opposite sides:
    • PQ = RS and QR = SP
  • Conclusion: Since opposite sides are equal, PQRS is a parallelogram.

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

Ans : 

Let the point P(x, y, z) be equidistant from the points A(1, 2, 3) and B(3, 2, -1).

Then, we can write:

PA = PB

Using the distance formula:

√((x – 1)^2 + (y – 2)^2 + (z – 3)^2) = √((x – 3)^2 + (y – 2)^2 + (z + 1)^2)

(x – 1)^2 + (y – 2)^2 + (z – 3)^2 = (x – 3)^2 + (y – 2)^2 + (z + 1)^2

Expanding both sides:

x^2 – 2x + 1 + y^2 – 4y + 4 + z^2 – 6z + 9 = x^2 – 6x + 9 + y^2 – 4y + 4 + z^2 + 2z + 1

Simplifying:

-2x + 14 = -6x + 14

Adding 6x and subtracting 14 from both sides:

4x = 0

Dividing both sides by 4:

x = 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.

Ans : 

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