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Introduction to Three-Dimensional Geometry - Section Formula

Grade 11ICSE

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

🔑Concepts

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The 3D Coordinate System: A point in three-dimensional space is represented as (x,y,z)(x, y, z). Visually, imagine three mutually perpendicular axes - the XX-axis (horizontal), YY-axis (vertical), and ZZ-axis (depth) - meeting at a central origin point O(0,0,0)O(0, 0, 0). These axes divide space into eight regions known as octants.

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Internal Division: When a point P(x,y,z)P(x, y, z) lies on the line segment joining two points A(x1,y1,z1)A(x_1, y_1, z_1) and B(x2,y2,z2)B(x_2, y_2, z_2) such that it divides the segment in the ratio m:nm:n internally. Visually, PP is located 'between' AA and BB, partitioning the distance into two parts proportional to mm and nn.

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External Division: When a point P(x,y,z)P(x, y, z) lies on the extension of the line segment joining A(x1,y1,z1)A(x_1, y_1, z_1) and B(x2,y2,z2)B(x_2, y_2, z_2) such that the ratio of distances AP:PBAP:PB is m:nm:n. Visually, PP is located outside the segment ABAB, either beyond BB (if m>nm > n) or before AA (if m<nm < n).

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The k:1k:1 Ratio Method: To find the ratio in which a point or a coordinate plane divides a line segment, it is convenient to assume the ratio as k:1k:1. If kk is positive, the division is internal; if kk is negative, the division is external. This reduces the number of unknowns from two (m,nm, n) to one (kk).

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Midpoint of a Segment: A special case of internal division where the ratio is 1:11:1. Visually, the midpoint is the exact geometric center of the line segment, equidistant from both endpoints AA and BB.

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Centroid of a Triangle: The point where the three medians of a triangle meet. In 3D space, if the vertices are A,B,A, B, and CC, the centroid GG is the average of the coordinates of the vertices. Visually, it represents the 'center of mass' of the triangular plate.

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Centroid of a Tetrahedron: A tetrahedron is a 3D solid with four triangular faces and four vertices. The centroid is the point where the lines joining each vertex to the centroid of the opposite face intersect. It is calculated by taking the arithmetic mean of the coordinates of all four vertices.

📐Formulae

Internal Section Formula: P(x,y,z)=(mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)P(x, y, z) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)

External Section Formula: P(x,y,z)=(mx2−nx1m−n,my2−ny1m−n,mz2−nz1m−n)P(x, y, z) = \left( \frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n} \right)

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

Centroid of a Triangle ABCABC: G(x,y,z)=(x1+x2+x33,y1+y2+y33,z1+z2+z33)G(x, y, z) = \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: G(x,y,z)=(x1+x2+x3+x44,y1+y2+y3+y44,z1+z2+z3+y44)G(x, y, z) = \left( \frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{y_1 + y_2 + y_3 + y_4}{4}, \frac{z_1 + z_2 + z_3 + y_4}{4} \right)

💡Examples

Problem 1:

Find the coordinates of the point which divides the line segment joining the points A(1,−2,3)A(1, -2, 3) and B(3,4,−5)B(3, 4, -5) in the ratio 2:32:3 internally.

Solution:

  1. Identify coordinates and ratio: (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, n=3n = 3.
  2. Apply the internal section formula for xx: x=2(3)+3(1)2+3=6+35=95x = \frac{2(3) + 3(1)}{2+3} = \frac{6+3}{5} = \frac{9}{5}.
  3. Apply the formula for yy: y=2(4)+3(−2)2+3=8−65=25y = \frac{2(4) + 3(-2)}{2+3} = \frac{8-6}{5} = \frac{2}{5}.
  4. Apply the formula for zz: z=2(−5)+3(3)2+3=−10+95=−15z = \frac{2(-5) + 3(3)}{2+3} = \frac{-10+9}{5} = \frac{-1}{5}.
  5. The required point is (95,25,−15)\left( \frac{9}{5}, \frac{2}{5}, \frac{-1}{5} \right).

Explanation:

This problem uses the internal section formula because the point is specified to divide the segment internally. We plug the given endpoints and ratio directly into the coordinates formula.

Problem 2:

Find the ratio in which the YZYZ-plane divides the line segment joining A(−2,4,7)A(-2, 4, 7) and B(3,−5,8)B(3, -5, 8).

Solution:

  1. Let the YZYZ-plane divide the segment ABAB in the ratio k:1k:1 at point PP.
  2. Any point on the YZYZ-plane has an xx-coordinate equal to 00. Therefore, xp=0x_p = 0.
  3. Using the section formula for the xx-coordinate: x=kx2+1x1k+1=0x = \frac{kx_2 + 1x_1}{k+1} = 0.
  4. Substitute the values: k(3)+1(−2)k+1=0\frac{k(3) + 1(-2)}{k+1} = 0.
  5. Solve for kk: 3k−2=0  ⟹  3k=2  ⟹  k=233k - 2 = 0 \implies 3k = 2 \implies k = \frac{2}{3}.
  6. Since kk is positive, the ratio is 2:32:3 internally.

Explanation:

When a plane divides a segment, we use the property of that specific plane (for YZYZ-plane, x=0x=0; for XZXZ-plane, y=0y=0; for XYXY-plane, z=0z=0). Setting the relevant coordinate to zero allows us to solve for the unknown ratio kk.