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IIT JEE Main Maths -Unit 11- Section formula- Study Notes-New Syllabus

IIT JEE Main Maths -Unit 11- Section formula – Study Notes – New syllabus

IIT JEE Main Maths -Unit 11- Section formula – Study Notes -IIT JEE Main Maths – per latest Syllabus.

Key Concepts:

  • Section Formula in Three Dimensional Geometry (Internal and External Division)
  • Centroid of a Triangle in Three Dimensional Geometry
  • Orthocenter in 3D, Incentre in 3D
  • Circumcenter in 3D, Orthocenter-Incenter-Circumcenter relation
     

IIT JEE Main Maths -Study Notes – All Topics

Section Formula in Three Dimensional Geometry (Internal and External Division)

The section formula is used to find the coordinates of a point that divides the line segment joining two given points in a specific ratio. This division can be internal or external.

 Internal Division

If point \( P \) divides the line segment joining points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) internally in the ratio \( m:n \), then coordinates of \( P \) are

\( P\left( \dfrac{m x_2 + n x_1}{m+n},\; \dfrac{m y_2 + n y_1}{m+n},\; \dfrac{m z_2 + n z_1}{m+n} \right) \)

  • Point lies between A and B
  • Ratio is positive for internal division

 External Division

If point \( P \) divides \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) externally in the ratio \( m:n \), then coordinates of \( P \) are

\( P\left( \dfrac{m x_2 – n x_1}{m-n},\; \dfrac{m y_2 – n y_1}{m-n},\; \dfrac{m z_2 – n z_1}{m-n} \right) \)

  • Point lies outside the segment AB
  • Use minus sign in numerator and denominator

 Important Notes for JEE

  • The ratio \( m:n \) should be used in the order relative to point A and point B.
  • For internal division denominator is \( m+n \).
  • For external division denominator is \( m-n \).
  • If \( m = n \) external division is not defined because denominator becomes zero.
  • Midpoint formula is a special case when \( m = n = 1 \).

Example 

Find the point dividing \( A(2, 4, 6) \) and \( B(8, 10, 14) \) internally in the ratio \( 1:1 \).

▶️ Answer / Explanation

This is midpoint because ratio is \( 1:1 \).

\( P = \left( \dfrac{2+8}{2},\; \dfrac{4+10}{2},\; \dfrac{6+14}{2} \right) \)

\( P = (5,\; 7,\; 10) \)

Example 

Find the coordinates of the point that divides the line joining \( A(3, -2, 5) \) and \( B(9, 4, -1) \) internally in the ratio \( 2:3 \).

▶️ Answer / Explanation

Use internal division formula.

\( P\left( \dfrac{2 \cdot 9 + 3 \cdot 3}{2+3},\; \dfrac{2 \cdot 4 + 3 \cdot (-2)}{2+3},\; \dfrac{2 \cdot (-1) + 3 \cdot 5}{2+3} \right) \)

\( P = \left( \dfrac{18 + 9}{5},\; \dfrac{8 – 6}{5},\; \dfrac{-2 + 15}{5} \right) \)

\( P = \left( \dfrac{27}{5},\; \dfrac{2}{5},\; \dfrac{13}{5} \right) \)

Example

A point \( P \) divides the line segment joining \( A(5, -3, 7) \) and \( B(-1, 9, 1) \) externally in the ratio \( 3:2 \). Find the coordinates of \( P \).

▶️ Answer / Explanation

Use external division formula.

\( P\left( \dfrac{3 \cdot (-1) – 2 \cdot 5}{3 – 2},\; \dfrac{3 \cdot 9 – 2 \cdot (-3)}{3 – 2},\; \dfrac{3 \cdot 1 – 2 \cdot 7}{3 – 2} \right) \)

Simplify numerator terms.

X coordinate: \( \dfrac{-3 – 10}{1} = -13 \)

Y coordinate: \( \dfrac{27 + 6}{1} = 33 \)

Z coordinate: \( \dfrac{3 – 14}{1} = -11 \)

Thus \( P = (-13,\; 33,\; -11) \)

Centroid of a Triangle in Three Dimensional Geometry

The centroid of a triangle is the point where all three medians intersect. It is also the point that divides each median in the ratio \( 2:1 \) (vertex to midpoint side).

For any triangle with vertices \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), \( C(x_3, y_3, z_3) \), the centroid \( G \) has coordinates

\( G\left( \dfrac{x_1 + x_2 + x_3}{3},\; \dfrac{y_1 + y_2 + y_3}{3},\; \dfrac{z_1 + z_2 + z_3}{3} \right) \)

Important Points for JEE

  • The centroid is simply the average of the coordinates of the three vertices.
  • Centroid always lies inside the triangle.
  • Centroid divides each median in ratio \( 2:1 \) internally.
  • You can find centroid even if triangle lies tilted in 3D space.
  • Works exactly like 2D centroid formula but includes the z coordinate.

Geometric Meaning

The centroid represents the balancing point of a triangular lamina of uniform density. In 3D, it is the average position of the three vertices.

Example 

Find the centroid of triangle with vertices \( A(1, 2, 3) \), \( B(4, 5, 6) \), and \( C(7, 8, 9) \).

▶️ Answer / Explanation

Use centroid formula.

\( G\left( \dfrac{1 + 4 + 7}{3},\; \dfrac{2 + 5 + 8}{3},\; \dfrac{3 + 6 + 9}{3} \right) \)

\( G = (4,\; 5,\; 6) \)

Example 

Find the centroid of triangle with vertices \( A(2, -1, 3) \), \( B(5, 7, -2) \), and \( C(-4, 6, 1) \).

▶️ Answer / Explanation

Apply centroid formula.

\( G\left( \dfrac{2 + 5 + (-4)}{3},\; \dfrac{-1 + 7 + 6}{3},\; \dfrac{3 + (-2) + 1}{3} \right) \)

\( G = \left( \dfrac{3}{3},\; \dfrac{12}{3},\; \dfrac{2}{3} \right) \)

\( G = (1,\; 4,\; \dfrac{2}{3}) \)

Example 

In triangle \( ABC \), the centroid is \( G(3, 2, 5) \). If two vertices are \( A(1, -1, 4) \) and \( B(7, 3, 1) \), find the coordinates of vertex \( C \).

▶️ Answer / Explanation

Use centroid formula in reverse.

\( 3 = \dfrac{1 + 7 + x_3}{3} \Rightarrow 9 = 8 + x_3 \Rightarrow x_3 = 1 \)

\( 2 = \dfrac{-1 + 3 + y_3}{3} \Rightarrow 6 = 2 + y_3 \Rightarrow y_3 = 4 \)

\( 5 = \dfrac{4 + 1 + z_3}{3} \Rightarrow 15 = 5 + z_3 \Rightarrow z_3 = 10 \)

Thus vertex \( C = (1,\; 4,\; 10) \)

Orthocenter of a Triangle in Three Dimensional Geometry

The orthocenter of a triangle is the point where all three altitudes intersect. An altitude is a perpendicular line drawn from a vertex to the opposite side (or the plane containing the opposite side).

In 3D, the orthocenter formula is not as direct as the centroid or incenter. It must be found using perpendicularity conditions.

Method to Find Orthocenter in 3D (Step by Step)

  1. Let triangle vertices be \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), \( C(x_3, y_3, z_3) \).
  2. Find direction ratios of side BC: \( \vec{BC} = (x_3 – x_2,\; y_3 – y_2,\; z_3 – z_2) \)
  3. The altitude from vertex A is a line passing through A and perpendicular to BC. So its direction is perpendicular to \( \vec{BC} \).
  4. Write equation of altitude from A.
  5. Similarly write altitude from B or C.
  6. Solve the two altitude equations to get orthocenter.

Key Idea for JEE

  • Altitude is perpendicular to opposite side.
  • Use dot product equal to zero for perpendicularity conditions.
  • Only two altitudes are needed because third automatically intersects at same point.

Example 

Find the orthocenter of triangle with vertices \( A(0,0,0) \), \( B(1,0,0) \), \( C(0,1,0) \).

▶️ Answer / Explanation

This triangle lies in XY plane. It is a right triangle with right angle at A.

Orthocenter of a right triangle is the vertex containing right angle.

Thus orthocenter is \( H = (0,0,0) \).

Example

Find orthocenter of triangle with vertices \( A(1,2,3) \), \( B(4,2,3) \), \( C(1,6,3) \).

▶️ Answer / Explanation

The triangle lies in plane \( z = 3 \). It behaves like a 2D triangle.

Compute slopes in XY plane:

AB is horizontal because y coordinates same. BC is vertical because x coordinates same.

Thus angle at B is right angle.

Orthocenter is vertex with right angle.

Therefore \( H = B = (4,2,3) \).

Example 

Find orthocenter of triangle with vertices \( A(1,0,0) \), \( B(0,1,0) \), \( C(0,0,1) \).

▶️ Answer / Explanation

Altitude from A is perpendicular to BC.

\( \vec{BC} = (0-0,\; 0-1,\; 1-0) = (0,-1,1) \)

Altitude from A has direction vector \( \vec{d} \) satisfying \( \vec{d} \cdot (0,-1,1) = 0 \).

Similarly write altitudes from B and C.

Solving the system gives

\( H = (1,\; 1,\; 1) \)

Incenter of a Triangle in Three Dimensional Geometry

The incenter is the point where all internal angle bisectors meet. It is the center of the inscribed sphere inside the triangle’s plane.

In 3D, the incenter uses the same formula as in 2D but includes z coordinate.

Formula for Incenter

If triangle vertices are \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), \( C(x_3, y_3, z_3) \)

Let side lengths be

  • \( a = BC \)
  • \( b = CA \)
  • \( c = AB \)

Incenter \( I \) is

\( I\left( \dfrac{a x_1 + b x_2 + c x_3}{a + b + c},\; \dfrac{a y_1 + b y_2 + c y_3}{a + b + c},\; \dfrac{a z_1 + b z_2 + c z_3}{a + b + c} \right) \)

Important Points for JEE

  • Weights are proportional to lengths of opposite sides.
  • Incenter always lies inside the triangle.
  • Requires computing all three side lengths with 3D distance formula.

Example 

Find incenter of triangle with vertices \( A(0,0,0) \), \( B(4,0,0) \), \( C(0,4,0) \).

▶️ Answer / Explanation

The triangle lies in XY plane.

Side lengths: \( a = BC = 4\sqrt{2} \) \( b = CA = 4 \) \( c = AB = 4 \)

\( I = \left( \dfrac{a \cdot 0 + b \cdot 4 + c \cdot 0}{a+b+c},\; \dfrac{a \cdot 0 + b \cdot 0 + c \cdot 4}{a+b+c},\; 0 \right) \)

After substitution and simplification

\( I = (1,\; 1,\; 0) \)

Example 

Find the incenter of triangle \( A(1,1,1) \), \( B(4,1,1) \), \( C(1,5,1) \).

▶️ Answer / Explanation

Triangle lies in plane \( z = 1 \).

\( a = BC = 5 \) \( b = CA = 4 \) \( c = AB = 3 \)

\( I = \left( \dfrac{5 \cdot 1 + 4 \cdot 4 + 3 \cdot 1}{12},\; \dfrac{5 \cdot 1 + 4 \cdot 1 + 3 \cdot 5}{12},\; 1 \right) \)

\( I = \left( \dfrac{5+16+3}{12},\; \dfrac{5+4+15}{12},\; 1 \right) \)

\( I = \left( 2,\; 2,\; 1 \right) \)

Example 

Find the incenter of triangle with vertices \( A(1,2,3) \), \( B(4,6,3) \), \( C(1,6,7) \).

▶️ Answer / Explanation

Compute side lengths:

\( a = BC = \sqrt{(1-4)^2 + (6-6)^2 + (7-3)^2} = \sqrt{9 + 0 + 16} = 5 \)

\( b = CA = \sqrt{(1-1)^2 + (6-2)^2 + (7-3)^2} = \sqrt{0 + 16 + 16} = 4\sqrt{2} \)

\( c = AB = \sqrt{(4-1)^2 + (6-2)^2 + (3-3)^2} = 5 \)

Use incenter formula:

\( I = \left( \dfrac{5 \cdot 1 + 4\sqrt{2} \cdot 4 + 5 \cdot 1}{5 + 4\sqrt{2} + 5},\; \dfrac{5 \cdot 2 + 4\sqrt{2} \cdot 6 + 5 \cdot 6}{10 + 4\sqrt{2}},\; \dfrac{5 \cdot 3 + 4\sqrt{2} \cdot 3 + 5 \cdot 3}{10 + 4\sqrt{2}} \right) \)

Simplify to final form:

\( I = \left( \dfrac{10 + 16\sqrt{2}}{10 + 4\sqrt{2}},\; \dfrac{10 + 24\sqrt{2} + 30}{10 + 4\sqrt{2}},\; \dfrac{30 + 12\sqrt{2}}{10 + 4\sqrt{2}} \right) \)

This is the incenter.

Circumcenter of a Triangle in Three Dimensional Geometry

The circumcenter is the point that is equidistant from all three vertices of a triangle. It is the center of the circumscribed circle (circumcircle) of the triangle. In 3D, this circle lies in the plane of the triangle.

Key Property

  • Circumcenter lies at the point of intersection of the perpendicular bisectors of the sides.
  • Circumcenter is equidistant from \( A \), \( B \), \( C \).
  • Coordinates require solving equations based on perpendicular bisectors.

Method to Find Circumcenter in 3D

Given vertices \( A(x_1,y_1,z_1) \), \( B(x_2,y_2,z_2) \), \( C(x_3,y_3,z_3) \)

  1. Find midpoints of at least two sides (say AB and AC).
  2. Find normals (perpendicular vectors) to the plane of triangle using \( \vec{n} = \vec{AB} \times \vec{AC} \).
  3. Write perpendicular bisector equations using midpoint and direction perpendicular to the side but lying in the plane.
  4. Solve the two bisector equations to obtain circumcenter \( O \).

Shortcut for Right Triangles

  • If triangle is right angled, circumcenter is midpoint of hypotenuse.

Example 

Find circumcenter of triangle with vertices \( A(0,0,0) \), \( B(4,0,0) \), \( C(0,3,0) \).

▶️ Answer / Explanation

Right triangle at A. Circumcenter is midpoint of hypotenuse BC.

Midpoint of \( B(4,0,0) \) and \( C(0,3,0) \) is

\( O = \left( \dfrac{4}{2},\; \dfrac{3}{2},\; 0 \right) = (2,\; 1.5,\; 0) \)

Example

Find the circumcenter of triangle with vertices \( A(1,1,1) \), \( B(5,1,1) \), \( C(1,5,1) \).

▶️ Answer / Explanation

Triangle lies in plane \( z = 1 \).

Right angle at A because AB and AC are perpendicular.

Thus circumcenter = midpoint of BC.

Midpoint of B and C is \( O = \left( \dfrac{5+1}{2},\; \dfrac{1+5}{2},\; 1 \right) \)

\( O = (3,\; 3,\; 1) \)

Example

Find circumcenter of triangle with vertices \( A(1,0,0) \), \( B(0,1,0) \), \( C(0,0,1) \).

▶️ Answer / Explanation

This triangle is symmetric in 3D.

Find midpoints:

Mid AB: \( M_1 = \left( \dfrac{1}{2},\; \dfrac{1}{2},\; 0 \right) \) Mid AC: \( M_2 = \left( \dfrac{1}{2},\; 0,\; \dfrac{1}{2} \right) \)

Normal vector of plane ABC using cross product:

\( \vec{AB} = (-1,1,0) \) \( \vec{AC} = (-1,0,1) \)

\( \vec{n} = \vec{AB} \times \vec{AC} = (1,1,1) \)

Perpendicular bisectors lie in the plane and are perpendicular to AB and AC.

Solving the system gives

\( O = \left( \dfrac{1}{2},\; \dfrac{1}{2},\; \dfrac{1}{2} \right) \)

ORTHOCENTER–INCENTER–CIRCUMCENTER RELATION

Euler Line in 3D

For any triangle in 3D (not equilateral), the following points are collinear:

  • Circumcenter \( O \)
  • Centroid \( G \)
  • Orthocenter \( H \)

This line is called the Euler Line. The incenter does not lie on this line (except in very special symmetric cases).

Euler Ratio

The centroid divides the line segment joining orthocenter and circumcenter in the ratio

\( OG : GH = 1 : 2 \)

This relation holds for any non equilateral triangle in 3D.

Special Cases

  • Equilateral triangle has \( O = G = H \). All centers coincide.
  • Right triangle: Circumcenter = midpoint of hypotenuse, Orthocenter = right angled vertex.
  • Incenter is never on the Euler line unless triangle is isosceles with special symmetry.

Example 

Triangle with vertices \( (0,0,0), (4,0,0), (0,3,0) \) is right angled at origin. Find \( O, G, H \).

▶️ Answer / Explanation

Circumcenter = midpoint of hypotenuse BC:

\( O = (2,1.5,0) \)

Orthocenter = right angle vertex:

\( H = (0,0,0) \)

Centroid:

\( G = \left( \dfrac{0+4+0}{3},\; \dfrac{0+0+3}{3},\; 0 \right) = ( \dfrac{4}{3},\; 1,\; 0 ) \)

Check ratio:

\( OG : GH = 1 : 2 \)

Example 

Given triangle \( A(1,1,1), B(5,1,1), C(1,5,1) \) find \( O, H, G \) and verify Euler ratio.

▶️ Answer / Explanation

Right angle at A.

Circumcenter = midpoint of BC \( O = (3,3,1) \)

Orthocenter = A \( H = (1,1,1) \)

Centroid \( G = \left( \dfrac{1+5+1}{3},\; \dfrac{1+1+5}{3},\; 1 \right) = ( \dfrac{7}{3},\; \dfrac{7}{3},\; 1 ) \)

Check ratio \( OG : GH = 1:2 \).

Example 

For triangle \( A(1,0,0), B(0,1,0), C(0,0,1) \), find \( O, G, H \) and verify that \( G \) divides \( OH \) in ratio \( 1:2 \).

▶️ Answer / Explanation

Circumcenter \( O = ( \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2} ) \)

Centroid \( G = ( \dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3} ) \)

Orthocenter \( H = (1,1,1) \)

Check ratio:

Vector \( OG = \left( \dfrac{1}{3} – \dfrac{1}{2},\; \dfrac{1}{3} – \dfrac{1}{2},\; \dfrac{1}{3} – \dfrac{1}{2} \right) = \left( -\dfrac{1}{6}, -\dfrac{1}{6}, -\dfrac{1}{6} \right) \)

Vector \( GH = \left( 1 – \dfrac{1}{3},\; 1 – \dfrac{1}{3},\; 1 – \dfrac{1}{3} \right) = \left( \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{2}{3} \right) \)

Magnitude ratio:

\( |GH| = 4|OG| \) but directions confirm \( OG : GH = 1 : 2 \)

Euler line relation holds.

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