Introduction

The section formula is a tool used to find the coordinates of a point that divides a line segment in a specific ratio. The formulas in 3D are a direct extension of the 2D versions.

1. Internal Division

If the point R divides the line segment joining points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2) internally in the ratio m : n, it means R lies between P and Q.

The coordinates of R are found by taking a weighted average:

R=(mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)R = \left( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n} \right)

2. External Division

If the point R divides the line segment joining P and Q externally in the ratio m : n, it means R lies on the line extended beyond P or Q.

The coordinates of R are found by changing the sign of 'n':

R=(mx2−nx1m−n,my2−ny1m−n,mz2−nz1m−n)R = \left( \frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}, \frac{mz_2-nz_1}{m-n} \right)

3. Midpoint Formula

The midpoint of the line segment is a special case of internal division where the ratio is 1:1 (m=n=1m=n=1).

M=(x1+x22,y1+y22,z1+z22)M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} \right)

4. Centroid of a Triangle and Tetrahedron

  • Centroid of a Triangle: The centroid is the point where the medians of a triangle intersect. Its coordinates are the average of the coordinates of the three vertices (x1,y1,z1)(x_1, y_1, z_1), (x2,y2,z2)(x_2, y_2, z_2), and (x3,y3,z3)(x_3, y_3, z_3):

G=(x1+x2+x33,y1+y2+y33,z1+z2+z33)G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)

  • Centroid of a Tetrahedron: Similarly, the centroid of a tetrahedron is the average of the coordinates of its four vertices.

Example 1: Internal Division

Question: Find the coordinates of the point which divides the line segment joining the points (1, –2, 3) and (3, 4, –5) in the ratio 2 : 3 internally.

Solution:

Here, (x1,y1,z1)=(1,−2,3)(x_1,y_1,z_1)=(1,-2,3), (x2,y2,z2)=(3,4,−5)(x_2,y_2,z_2)=(3,4,-5), m=2m=2, and n=3n=3. We use the internal section formula for each coordinate:

  • x=mx2+nx1m+n=2(3)+3(1)2+3=6+35=95x = \frac{mx_2+nx_1}{m+n} = \frac{2(3)+3(1)}{2+3} = \frac{6+3}{5} = \frac{9}{5}

  • y=my2+ny1m+n=2(4)+3(−2)2+3=8−65=25y = \frac{my_2+ny_1}{m+n} = \frac{2(4)+3(-2)}{2+3} = \frac{8-6}{5} = \frac{2}{5}

  • z=mz2+nz1m+n=2(−5)+3(3)2+3=−10+95=−15z = \frac{mz_2+nz_1}{m+n} = \frac{2(-5)+3(3)}{2+3} = \frac{-10+9}{5} = -\frac{1}{5}

The point is (9/5, 2/5, -1/5).

Example 2: External Division

Question: Find the coordinates of the point which divides the line segment joining P(2,-1,4) and Q(3,1,5) in the ratio 2:1 externally.

Solution:

Here, (x1,y1,z1)=(2,−1,4)(x_1,y_1,z_1)=(2,-1,4), (x2,y2,z2)=(3,1,5)(x_2,y_2,z_2)=(3,1,5), m=2m=2, and n=1n=1. We use the external section formula:

  • x=mx2−nx1m−n=2(3)−1(2)2−1=41=4x = \frac{mx_2-nx_1}{m-n} = \frac{2(3)-1(2)}{2-1} = \frac{4}{1} = 4

  • y=my2−ny1m−n=2(1)−1(−1)2−1=2+11=3y = \frac{my_2-ny_1}{m-n} = \frac{2(1)-1(-1)}{2-1} = \frac{2+1}{1} = 3

  • z=mz2−nz1m−n=2(5)−1(4)2−1=10−41=6z = \frac{mz_2-nz_1}{m-n} = \frac{2(5)-1(4)}{2-1} = \frac{10-4}{1} = 6

The point is (4,3,6).

Example 3: Finding the Ratio of Division

Question: Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, –8) is divided by the YZ-plane.

Solution:

Step 1: Any point on the YZ-plane has its x-coordinate equal to 0.

Step 2: Let the ratio be k:1. We use the section formula for the x-coordinate:

x=k(x2)+1(x1)k+1=k(6)+1(4)k+1x = \frac{k(x_2)+1(x_1)}{k+1} = \frac{k(6)+1(4)}{k+1}

Step 3: Set the x-coordinate to 0 and solve for k.

6k+4k+1=0  ⟹  6k+4=0  ⟹  k=−4/6=−2/3\frac{6k+4}{k+1} = 0 \implies 6k+4 = 0 \implies k = -4/6 = -2/3

Step 4: Interpret the ratio. Since k is negative, the division is external. The ratio is ∣k∣=2/3|k|=2/3, so the YZ-plane divides the segment in the ratio 2:3 externally.

Example 4: Collinearity using Section Formula

Question: Using the section formula, show that the points A(2, –3, 4), B(–1, 2, 1) and C(0, 1/3, 2) are collinear.

Solution:

If the points are collinear, then one point must divide the line segment formed by the other two in some ratio. Let's assume C divides AB in the ratio k:1.

Step 1: Use the x-coordinate to find the potential ratio k.

xC=k(xB)+1(xA)k+1  ⟹  0=k(−1)+1(2)k+1  ⟹  0=−k+2  ⟹  k=2x_C = \frac{k(x_B)+1(x_A)}{k+1} \implies 0 = \frac{k(-1)+1(2)}{k+1} \implies 0 = -k+2 \implies k=2

Step 2: Verify this ratio for the y and z coordinates.

  • y=2(yB)+1(yA)2+1=2(2)+1(−3)3=4−33=13y = \frac{2(y_B)+1(y_A)}{2+1} = \frac{2(2)+1(-3)}{3} = \frac{4-3}{3} = \frac{1}{3} (Matches yCy_C)

  • z=2(zB)+1(zA)2+1=2(1)+1(4)3=2+43=2z = \frac{2(z_B)+1(z_A)}{2+1} = \frac{2(1)+1(4)}{3} = \frac{2+4}{3} = 2 (Matches zCz_C)

Since the point of division for k=2 is (0, 1/3, 2), which is exactly point C, we can conclude that C divides AB internally in the ratio 2:1. Therefore, the points are collinear.

Example 5: Finding a Vertex from the Centroid

Question: If the origin O(0,0,0) is the centroid of a triangle with vertices A(a,1,3), B(-2,b,-5) and C(4,7,c), find the values of a, b, and c.

Solution:

The coordinates of the centroid are the average of the coordinates of the vertices. We set up an equation for each coordinate.

  • x-coordinate: a+(−2)+43=0  ⟹  a+2=0  ⟹  a=−2\frac{a+(-2)+4}{3} = 0 \implies a+2=0 \implies \mathbf{a=-2}

  • y-coordinate: 1+b+73=0  ⟹  b+8=0  ⟹  b=−8\frac{1+b+7}{3} = 0 \implies b+8=0 \implies \mathbf{b=-8}

  • z-coordinate: 3+(−5)+c3=0  ⟹  c−2=0  ⟹  c=2\frac{3+(-5)+c}{3} = 0 \implies c-2=0 \implies \mathbf{c=2}