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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 Section Formula (Internal Division) determines the coordinates of a point PP that divides 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) in a given ratio m:nm : n. The point PP lies on the segment ABAB such that APPB=mn\frac{AP}{PB} = \frac{m}{n}.

A line segment AB divided internally by point P in ratio m:n
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The External Division occurs when a point PP lies on the extension of the line segment ABAB such that AP:BP=m:nAP : BP = m : n. In this case, the formula uses a negative sign between the ratio terms in the numerator and denominator.

A line segment AB extended to point P for external division
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The Midpoint of a line segment is a special case of the section formula where the ratio m:nm : n is 1:11 : 1. It represents the average of the xx, yy, and zz coordinates of the endpoints.

Line segment AB with M as the midpoint
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The Centroid of a Triangle (GG) is the point where the medians intersect. It divides each median in the ratio 2:12 : 1 from the vertex to the midpoint of the opposite side.

Centroid G of a triangle ABC

📐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.

Problem 3:

Determine the coordinates of the point QQ which divides the line segment joining L(2,4,5)L(2, 4, 5) and M(3,5,−4)M(3, 5, -4) externally in the ratio 3:23:2.

Point Q dividing LM externally in ratio 3:2

Solution:

Given: A(x1,y1,z1)=(2,4,5)A(x_1, y_1, z_1) = (2, 4, 5), B(x2,y2,z2)=(3,5,−4)B(x_2, y_2, z_2) = (3, 5, -4), and ratio m:n=3:2m:n = 3:2. Using the external section formula: x=mx2−nx1m−n=3(3)−2(2)3−2=9−41=5x = \frac{m x_2 - n x_1}{m - n} = \frac{3(3) - 2(2)}{3 - 2} = \frac{9 - 4}{1} = 5 y=my2−ny1m−n=3(5)−2(4)3−2=15−81=7y = \frac{m y_2 - n y_1}{m - n} = \frac{3(5) - 2(4)}{3 - 2} = \frac{15 - 8}{1} = 7 z=mz2−nz1m−n=3(−4)−2(5)3−2=−12−101=−22z = \frac{m z_2 - n z_1}{m - n} = \frac{3(-4) - 2(5)}{3 - 2} = \frac{-12 - 10}{1} = -22 Therefore, the coordinates of QQ are (5,7,−22)(5, 7, -22).

Explanation:

To divide a segment externally, the ratio m:nm:n is applied with a subtraction in the formula. Point QQ lies on the line passing through LL and MM, but outside the segment LMLM, closer to MM because m>nm > n.

Problem 4:

Find the ratio in which the XZXZ-plane divides the line segment joining the points P(−3,4,−8)P(-3, 4, -8) and Q(5,−6,4)Q(5, -6, 4). Also, find the coordinates of the point of intersection.

Line PQ intersecting the XZ-plane at point R

Solution:

Let the XZXZ-plane divide PQPQ in the ratio k:1k : 1 at point R(x,y,z)R(x, y, z). In the XZXZ-plane, the yy-coordinate is always 00. Using the section formula for the yy-coordinate: y=ky2+1y1k+1=0y = \frac{k y_2 + 1 y_1}{k + 1} = 0 k(−6)+1(4)k+1=0\frac{k(-6) + 1(4)}{k + 1} = 0 −6k+4=0  ⟹  6k=4  ⟹  k=23-6k + 4 = 0 \implies 6k = 4 \implies k = \frac{2}{3} The ratio is 2:32:3 internally. Now, find xx and zz: x=23(5)+1(−3)23+1=103−353=1/35/3=15x = \frac{\frac{2}{3}(5) + 1(-3)}{\frac{2}{3} + 1} = \frac{\frac{10}{3} - 3}{\frac{5}{3}} = \frac{1/3}{5/3} = \frac{1}{5} z=23(4)+1(−8)23+1=83−853=−16/35/3=−165z = \frac{\frac{2}{3}(4) + 1(-8)}{\frac{2}{3} + 1} = \frac{\frac{8}{3} - 8}{\frac{5}{3}} = \frac{-16/3}{5/3} = -\frac{16}{5} The point of intersection is (15,0,−165)(\frac{1}{5}, 0, -\frac{16}{5}).

Explanation:

Since the XZXZ-plane is defined by y=0y=0, we set the yy-coordinate of the dividing point to zero to find the ratio kk. Once kk is found, we substitute it back to find the remaining coordinates.