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Introduction to Three Dimensional Geometry - Coordinate Axes and Coordinate Planes in Three Dimensional Space

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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In three-dimensional geometry, we use three mutually perpendicular lines passing through a common point OO (the origin). These lines are called the XX, YY, and ZZ axes. The position of any point PP is given by an ordered triplet (x,y,z)(x, y, z). The space is divided into 8 regions called octants.

Three-dimensional coordinate system showing X, Y, and Z axes intersecting at the origin O.
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The three pairs of axes define three coordinate planes: the XYXY-plane (where z=0z=0), the YZYZ-plane (where x=0x=0), and the ZXZX-plane (where y=0y=0). Every point in space lies in relation to these planes.

A representation of the XY plane in 3D space where the z-coordinate is zero.
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The signs of the coordinates (x,y,z)(x, y, z) determine the octant. For example, in the first octant (I), all coordinates are positive (+,+,+)(+, +, +). In the fifth octant (V), xx and yy are positive while zz is negative (+,+,−)(+, +, -).

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The distance of a point P(x,y,z)P(x, y, z) from the XYXY-plane is ∣z∣|z|, from the YZYZ-plane is ∣x∣|x|, and from the ZXZX-plane is ∣y∣|y|.

📐Formulae

Coordinates of the origin: (0,0,0)(0, 0, 0) contrast to 2D (0,0)(0, 0)

Equation of the XYXY-plane: z=0z = 0

Equation of the YZYZ-plane: x=0x = 0

Equation of the ZXZX-plane: y=0y = 0

Equation of the XX-axis: y=0,z=0y = 0, z = 0

Equation of the YY-axis: x=0,z=0x = 0, z = 0

Equation of the ZZ-axis: x=0,y=0x = 0, y = 0

Distance of a point P(x,y,z)P(x, y, z) from the origin O(0,0,0)O(0, 0, 0): OP=x2+y2+z2OP = \sqrt{x^2 + y^2 + z^2}

Distance between two points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2): PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

💡Examples

Problem 1:

Determine the octant in which the following points lie: (i) (4,−2,3)(4, -2, 3) and (ii) (−4,2,−5)(-4, 2, -5).

Solution:

For point (i) (4,−2,3)(4, -2, 3), we have x>0x > 0 (positive), y<0y < 0 (negative), and z>0z > 0 (positive). This corresponds to Octant IV (XOY′ZXOY'Z). For point (ii) (−4,2,−5)(-4, 2, -5), we have x<0x < 0 (negative), y>0y > 0 (positive), and z<0z < 0 (negative). This corresponds to Octant VI (X′OYZ′X'OYZ').

Explanation:

To identify the octant, check the signs of the x,y,x, y, and zz coordinates. There are 8 combinations: (+,+,+) is I, (-,+,+) is II, (-,-,+) is III, (+,-,+) is IV. Adding a negative zz maps to V, VI, VII, VIII respectively.

Problem 2:

A point PP is at a distance of 55 units from the XX-axis and lies on the YY-axis. What are its coordinates?

Solution:

Since the point lies on the YY-axis, its xx and zz coordinates must be zero. Thus, the point is of the form (0,y,0)(0, y, 0). The distance from the XX-axis to a point (x,y,z)(x, y, z) is given by y2+z2\sqrt{y^2 + z^2}. Here, y2+02=5\sqrt{y^2 + 0^2} = 5, which means ∣y∣=5|y| = 5. Therefore, the coordinates are (0,5,0)(0, 5, 0) or (0,−5,0)(0, -5, 0).

Explanation:

We use the property that points on the YY-axis have x=0x=0 and z=0z=0, and then apply the geometric definition of distance from an axis.

Problem 3:

Find the coordinates of the foot of the perpendicular drawn from the point A(3,4,5)A(3, 4, 5) to the ZXZX-plane.

Diagram showing point A projected onto the ZX plane at point M.

Solution:

  1. Any point on the ZXZX-plane has its yy-coordinate equal to 00.
  2. The perpendicular from A(3,4,5)A(3, 4, 5) to the ZXZX-plane keeps the xx and zz coordinates same but reduces the distance from the plane to zero.
  3. Therefore, the foot of the perpendicular is M(3,0,5)M(3, 0, 5).

Explanation:

In 3D space, projecting a point onto a coordinate plane involves setting the coordinate corresponding to the 'missing' axis in the plane's name to zero.

Problem 4:

A point QQ lies on the ZZ-axis. If its distance from the point (0,0,0)(0, 0, 0) is 77 units and it lies in the negative direction of the ZZ-axis, what are its coordinates?

Coordinate axis showing point Q located at -7 on the Z-axis.

Solution:

  1. Any point on the ZZ-axis has coordinates in the form (0,0,z)(0, 0, z).
  2. The distance from the origin is given as 77 units, so ∣z∣=7|z| = 7.
  3. Since the point lies in the negative direction of the ZZ-axis, z=−7z = -7.
  4. Thus, the coordinates are (0,0,−7)(0, 0, -7).

Explanation:

Points on an axis have two of their coordinates as zero. The sign is determined by the direction (positive or negative) along that axis.