Space Coordinates

Just as a point in the plane is pinned down by two numbers (x, y), a point in space is located by three numbers (x, y, z). Adding just one more number for height, z, is like lifting a point off the floor and up into the air. The distance between two points and the equation of a sphere both come straight from their flat, 2D versions: you simply add one more layer of z. Drag the x, y, and z sliders to see where the point sits in space and the shadow (projection) it casts onto the coordinate plane, and use the radius r slider to watch the sphere grow and shrink.

3D Coordinate System

x-coordinate3

y-coordinate4

z-coordinate2

👀 Points in Space

①Point P(x,y,z) is determined by three coordinates

②Dashed lines are projections onto each axis and coordinate plane

③Shadow on xy-plane: (x,y,0)

④Shadow on z-axis: (0,0,z)

Distance Between Two Points

Distance Between Two Points in Space

d = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²)

Pythagorean theorem extended to 3D

📐 Intuition for the Distance Formula

①2D distance formula: √(Δx² + Δy²)

②For 3D just add one Δz²

③Trace the right-angle path drawn with dashed lines

④Apply right triangle twice in order Δx → Δy → Δz

Equation of a Sphere

Radius r3

Equation of a Sphere

(x−a)² + (y−b)² + (z−c)² = r²

Center C(a,b,c), radius r

🌐 What is a Sphere?

①Set of points equidistant (radius) from a fixed point (center)

②3D extension of the circle equation

③Expanded form: x² + y² + z² + Dx + Ey + Fz + G = 0

④Complete the square in the general form to find center and radius

Internal Division and Centroid

Internal Division (3D)

P = (nx₁+mx₂m+n, ny₁+my₂m+n, nz₁+mz₂m+n)

Point dividing segment AB in ratio m:n

Centroid of a Triangle

G = (x₁+x₂+x₃3, y₁+y₂+y₃3, z₁+z₂+z₃3)

Average of the three vertex coordinates

💡 Mnemonic for Division Point

①Same structure as the 2D division formula

②Just add one z-coordinate

③Midpoint is the special case m=n=1: average of coordinates

④Centroid: average of all three coordinates

Work It Out

Example 1

Find the distance between the two points A(1, 2, 3) and B(4, 6, 3).

Square the difference of each coordinate and add them.

AB = √((4−1)² + (6−2)² + (3−3)²)

Compute the sum of squares and take the square root.

AB = √(9 + 16 + 0) = √25 = 5

▸ 5

When the z-coordinates match, Δz=0 and it reduces to the 2D distance formula.

Example 2

Find the center and radius of the sphere x²+y²+z²−2x+4y−6z+5=0.

Complete the square in x, y, and z.

x²−2x = (x−1)²−1, y²+4y = (y+2)²−4, z²−6z = (z−3)²−9

Compare with the standard form to read off center and radius.

(x−1)² + (y+2)² + (z−3)² = 1+4+9−5 = 9

▸ center (1, −2, 3), radius 3

The right side is r², so √9=3 is the radius. A negative right side means no sphere exists.

Summary

Distance from Origin

d = √(x² + y² + z²)

Distance from origin O to point P(x,y,z)

2023 KICE mock exam Math (Geometry) type, adapted

What is the center of the sphere whose diameter endpoints are A(2, 0, 1) and B(0, 4, 3)?

①(1, 2, 2)

②(2, 2, 1)

③(1, 2, 1)

④(2, 4, 2)

⑤(1, 4, 2)

▸ ① (1, 2, 2)

The midpoint of the diameter endpoints is the center of the sphere.

Average each coordinate.

((2+0)/2, (0+4)/2, (1+3)/2) = (1, 2, 2)

🎯 Exam Points

①Space distance formula: just add Δz² to 2D

②Sphere equation: add z-term to circle equation

③Convert general → standard form to find center and radius

④Division and centroid: same structure as 2D, add z

⑤Practice symmetry points: about axes, planes, and points

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