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Introduction to Three-Dimensional Geometry - Coordinate Axes and Coordinate Planes in 3D

Grade 11ICSE

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

🔑Concepts

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The 3D coordinate system consists of three mutually perpendicular lines passing through a fixed point called the origin O(0,0,0)O(0, 0, 0). These lines are the XX-axis, YY-axis, and ZZ-axis.

3D coordinate axes showing X, Y, and Z axes intersecting at origin O.
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Three coordinate planes are formed by pairs of axes: the XYXY-plane (contains XX and YY axes, z=0z=0), the YZYZ-plane (contains YY and ZZ axes, x=0x=0), and the ZXZX-plane (contains ZZ and XX axes, y=0y=0).

Illustration of the XY coordinate plane.
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The coordinate planes divide the space into eight regions called octants. The sign of coordinates (x,y,z)(x, y, z) determines the octant. For example, in the first octant, all coordinates are positive.

Representation of Octant I where x, y, and z are all positive.
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Any point PP in space is represented as (x,y,z)(x, y, z), where x,y,x, y, and zz are the perpendicular distances from the YZYZ, ZXZX, and XYXY planes respectively.

📐Formulae

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

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

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

Coordinates of a point on the XX-axis: (x,0,0)(x, 0, 0)

Coordinates of a point on the YY-axis: (0,y,0)(0, y, 0)

Coordinates of a point on the ZZ-axis: (0,0,z)(0, 0, z)

Distance of point P(x,y,z)P(x, y, z) from the origin: d=sqrtx2+y2+z2d = \\sqrt{x^2 + y^2 + z^2}

Distance between two points P1(x1,y1,z1)P_1(x_1, y_1, z_1) and P2(x2,y2,z2)P_2(x_2, y_2, z_2): d=sqrt(x2−x1)2+(y2−y1)2+(z2−z1)2d = \\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

💡Examples

Problem 1:

Determine the octants in which the following points lie: A(2,−3,5)A(2, -3, 5) and B(−1,−4,−2)B(-1, -4, -2).

Solution:

  1. For point A(2,−3,5)A(2, -3, 5): Here xx is positive (++), yy is negative (−-), and zz is positive (++). Looking at the sign convention, the (+,−,+)(+, -, +) pattern corresponds to Octant IV.
  2. For point B(−1,−4,−2)B(-1, -4, -2): Here xx is negative (−-), yy is negative (−-), and zz is negative (−-). The (−,−,−)(- , - , -) pattern corresponds to Octant VII.

Explanation:

To identify the octant, we look at the signs of the x,y,zx, y, z coordinates. Octants I-IV have z>0z > 0 and Octants V-VIII have z<0z < 0.

Problem 2:

Find the distance of the point P(3,−4,12)P(3, -4, 12) from (i) the XYXY-plane and (ii) the Origin.

Solution:

  1. Distance from the XYXY-plane: The perpendicular distance of any point (x,y,z)(x, y, z) from the XYXY-plane is given by ∣z∣|z|. Here, z=12z = 12, so the distance is ∣12∣=12|12| = 12 units.
  2. Distance from the Origin O(0,0,0)O(0, 0, 0): Using the distance formula d=sqrtx2+y2+z2d = \\sqrt{x^2 + y^2 + z^2}, we get: d=sqrt32+(−4)2+122d = \\sqrt{3^2 + (-4)^2 + 12^2} d=sqrt9+16+144d = \\sqrt{9 + 16 + 144} d=sqrt169=13d = \\sqrt{169} = 13 units.

Explanation:

The distance from a coordinate plane is the absolute value of the 'missing' coordinate. The distance from the origin uses the 3D version of the Pythagorean theorem.

Problem 3:

Find the coordinates of the point which is the projection of A(4,5,6)A(4, 5, 6) on the XYXY-plane, YZYZ-plane, and ZXZX-plane.

Diagram showing projection of point A onto the XY plane.

Solution:

  1. For the XYXY-plane, the zz-coordinate becomes zero. So, projection is P1(4,5,0)P_1(4, 5, 0).
  2. For the YZYZ-plane, the xx-coordinate becomes zero. So, projection is P2(0,5,6)P_2(0, 5, 6).
  3. For the ZXZX-plane, the yy-coordinate becomes zero. So, projection is P3(4,0,6)P_3(4, 0, 6).

Explanation:

The projection of a point (x,y,z)(x, y, z) onto a coordinate plane is found by setting the coordinate corresponding to the missing axis of that plane to zero.

Problem 4:

Show that the points L(0,7,10)L(0, 7, 10), M(−1,6,6)M(-1, 6, 6) and N(−4,9,6)N(-4, 9, 6) form a right-angled triangle.

Right-angled triangle LMN with right angle at vertex M.

Solution:

Using distance formula d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}: LM2=(−1−0)2+(6−7)2+(6−10)2=1+1+16=18LM^2 = (-1-0)^2 + (6-7)^2 + (6-10)^2 = 1 + 1 + 16 = 18 MN2=(−4+1)2+(9−6)2+(6−6)2=9+9+0=18MN^2 = (-4+1)^2 + (9-6)^2 + (6-6)^2 = 9 + 9 + 0 = 18 LN2=(−4−0)2+(9−7)2+(6−10)2=16+4+16=36LN^2 = (-4-0)^2 + (9-7)^2 + (6-10)^2 = 16 + 4 + 16 = 36 Since LM2+MN2=18+18=36=LN2LM^2 + MN^2 = 18 + 18 = 36 = LN^2, the triangle is right-angled at MM.

Explanation:

We calculate the squares of the lengths of the three sides using the 3D distance formula and verify if the Pythagorean theorem a2+b2=c2a^2 + b^2 = c^2 holds true.

Coordinate Axes and Coordinate Planes in 3D Class 11 Notes & Examples