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Introduction to Three Dimensional Geometry - Section Formula

Grade 11CBSE

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

🔑Concepts

Position of a Point in 3D Space: A point in three-dimensional space is represented as (x,y,z)(x, y, z), which describes its distances from the YZYZ-plane, ZXZX-plane, and XYXY-plane respectively. Visually, think of a point PP as a vertex of a rectangular box whose three adjacent edges lie along the XX, YY, and ZZ axes.

Internal Division Concept: When a point RR lies on the line segment joining 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) such that it divides the segment in a given ratio m:nm:n internally. Visually, RR is located between PP and QQ, and the ratio of the lengths PRPR to RQRQ is exactly m/nm/n.

External Division Concept: When a point RR lies on the line passing through P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2) but outside the segment PQPQ, it is said to divide PQPQ externally in the ratio m:nm:n. Visually, RR is placed on the extension of the line segment PQPQ, where the distance from PP to RR and QQ to RR maintains the ratio m:nm:n.

Midpoint as a Special Case: The midpoint is a specific instance of internal division where the ratio m:nm:n is 1:11:1. Visually, the midpoint MM is the unique point on the segment PQPQ that is equidistant from both endpoints PP and QQ.

Centroid of a Triangle: In 3D geometry, the centroid of a triangle with 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) is the point where the three medians intersect. Visually, it represents the geometric center or the 'balance point' of the triangular surface in space.

The Ratio k:1k:1 Method: To find the ratio in which a point divides a line segment, it is often mathematically simpler to assume the ratio is k:1k:1. If the resulting value of kk is positive, the division is internal; if kk is negative, the division is external.

📐Formulae

Internal Division: 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)

External Division: R=(mx2nx1mn,my2ny1mn,mz2nz1mn)R = \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=(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)

Centroid of a Triangle: 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)

Section Formula using k:1k:1 ratio: R=(kx2+x1k+1,ky2+y1k+1,kz2+z1k+1)R = \left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1}, \frac{kz_2 + z_1}{k+1} \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:

Given: A(x1,y1,z1)=(1,2,3)A(x_1, y_1, z_1) = (1, -2, 3), B(x2,y2,z2)=(3,4,5)B(x_2, y_2, z_2) = (3, 4, -5), m=2m = 2, and n=3n = 3. \ Using the internal section formula: \ x=mx2+nx1m+n=2(3)+3(1)2+3=6+35=95x = \frac{m x_2 + n x_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=865=25y = \frac{m y_2 + n y_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{m z_2 + n z_1}{m+n} = \frac{2(-5) + 3(3)}{2+3} = \frac{-10+9}{5} = -\frac{1}{5} \ Therefore, the coordinates are (95,25,15)(\frac{9}{5}, \frac{2}{5}, -\frac{1}{5}).

Explanation:

We apply the internal division formula by substituting the coordinates of the given points and the values of the ratio mm and nn into the respective xx, yy, and zz components.

Problem 2:

Find the ratio in which the YZYZ-plane divides the line segment formed by joining the points (2,4,7)(-2, 4, 7) and (3,5,8)(3, -5, 8).

Solution:

Let the YZYZ-plane divide the line segment joining A(2,4,7)A(-2, 4, 7) and B(3,5,8)B(3, -5, 8) in the ratio k:1k:1 at point PP. \ On the YZYZ-plane, the xx-coordinate of any point is always zero. \ The xx-coordinate of point PP is given by: \ x=k(3)+1(2)k+1x = \frac{k(3) + 1(-2)}{k+1} \ Since x=0x = 0 on the YZYZ-plane: \ 0=3k2k+10 = \frac{3k - 2}{k+1} \ 3k2=03k - 2 = 0 \ 3k=2    k=233k = 2 \implies k = \frac{2}{3} \ Thus, the ratio is 2:32:3 internally.

Explanation:

To find the ratio, we use the property that the xx-coordinate is zero on the YZYZ-plane. We assume the ratio is k:1k:1, set the xx-component of the section formula to zero, and solve for kk.