Chapter 11 — INTRODUCTION TO THREE DIMENSIONAL GEOMETRY

Theory Used

In a parallelogram, the diagonals bisect each other. This implies that the midpoint of one diagonal is equal to the midpoint of the other diagonal.

If points A, B, C, D form a parallelogram, then: \[ \text{Midpoint of } AC = \text{Midpoint of } BD \]

Solution Roadmap

• Compute midpoint of diagonal AC
• Assume coordinates of D = (x, y, z)
• Apply midpoint formula on BD
• Equate both midpoints and solve for x, y, z

Solution

Given: \[ A(3, -1, 2), \quad B(1, 2, -4), \quad C(-1, 1, 2) \]

Midpoint of AC: \[ \begin{aligned} E_x &= \frac{3 + (-1)}{2} = 1 \\ E_y &= \frac{-1 + 1}{2} = 0 \\ E_z &= \frac{2 + 2}{2} = 2 \end{aligned} \] Hence, midpoint \(E = (1, 0, 2)\)

Let \(D = (x, y, z)\)

Midpoint of BD: \[ \left( \frac{1 + x}{2}, \frac{2 + y}{2}, \frac{-4 + z}{2} \right) \]

Equating with midpoint E: \[ \begin{aligned} \frac{1 + x}{2} &= 1 \Rightarrow x = 1 \\ \frac{2 + y}{2} &= 0 \Rightarrow y = -2 \\ \frac{-4 + z}{2} &= 2 \Rightarrow z = 8 \end{aligned} \]

Therefore, the fourth vertex is: \[ D(1, -2, 8) \]

Geometric Illustration

Why This Question Matters

• Direct application of midpoint formula — frequently tested in CBSE board exams
• Builds foundation for vector geometry and 3D coordinate algebra
• Common pattern in JEE Main (coordinate + vector hybrid questions)
• Enhances spatial reasoning using algebraic symmetry

Theory Used

A median of a triangle is the line segment joining a vertex to the midpoint of the opposite side.

Key formulas used: \[ \text{Midpoint} = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right) \] \[ \text{Distance} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \]

Solution Roadmap

• Find midpoint of each side of the triangle
• Join each vertex to midpoint of opposite side
• Apply distance formula to compute median lengths
• Compare results for symmetry insight

Solution

Given: \[ A(0,0,6), \quad B(0,4,0), \quad C(6,0,0) \]

Midpoint of \(BC\): \[ (3,2,0) \]

Length of median from \(A\): \[ \begin{aligned} \sqrt{(3-0)^2+(2-0)^2+(0-6)^2} &= \sqrt{49} \\&= 7 \end{aligned} \]

Midpoint of \(AB\): \[ (0,2,3) \]

Length of median from \(C\): \[ \begin{aligned} \sqrt{(0-6)^2+(2-0)^2+(3-0)^2} &= \sqrt{49} \\&= 7 \end{aligned} \]

Midpoint of \(AC\): \[ (3,0,3) \]

Length of median from \(B\): \[ \sqrt{(3-0)^2+(0-4)^2+(3-0)^2} = \sqrt{34} \]

Final Answer: \[ 7,\; 7,\; \sqrt{34} \]

Geometric Insight (Minimal 3D Projection)

Why This Question Matters

• Direct application of midpoint + distance formula — high probability board question
• Tests multi-step coordinate handling accuracy (common mistake zone)
• Important for JEE Main vector geometry problems involving centroids and medians
• Builds intuition for symmetry in 3D coordinate systems

Exam Insight

Two medians being equal indicates partial symmetry in the triangle. In competitive exams, such patterns often hint toward isosceles structures or vector simplifications.

Theory Used

The centroid of a triangle is the average of the coordinates of its vertices.

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

If the centroid is the origin, then each coordinate sum must be zero: \[ x_1+x_2+x_3 = 0,\quad y_1+y_2+y_3 = 0,\quad z_1+z_2+z_3 = 0 \]

Solution Roadmap

• Apply centroid condition coordinate-wise
• Form three linear equations in a, b, c
• Solve each independently
• Verify consistency

Solution

Given vertices: \[ P(2a,2,6),\quad Q(-4,3b,-10),\quad R(8,14,2c) \]

Since centroid is origin:

x-coordinate condition: \[ \frac{2a - 4 + 8}{3} = 0 \Rightarrow 2a + 4 = 0 \Rightarrow a = -2 \]

y-coordinate condition: \[ \frac{2 + 3b + 14}{3} = 0 \Rightarrow 3b + 16 = 0 \Rightarrow b = -\frac{16}{3} \]

z-coordinate condition: \[ \frac{6 - 10 + 2c}{3} = 0 \Rightarrow 2c - 4 = 0 \Rightarrow c = 2 \]

Final Answer: \[ a = -2,\quad b = -\frac{16}{3},\quad c = 2 \]

Geometric Insight

Why This Question Matters

• Centroid-based equations are a standard CBSE board pattern
• Tests ability to translate geometry into algebraic constraints
• Frequently appears in JEE Main as a direct formula application
• Builds foundation for vector barycentric concepts

Exam Insight

Shortcut: If centroid is origin, directly equate sum of coordinates to zero instead of writing full formula. This reduces computation time significantly in competitive exams.

Theory Used

In 3D geometry, the locus defined by the sum of squares of distances from two fixed points is a sphere.

Key identity: \[ PA^2 + PB^2 = 2\left( PM^2 + \frac{AB^2}{4} \right) \] where \(M\) is the midpoint of \(AB\).

This transforms the locus into a standard sphere equation centered at midpoint of A and B.

Solution Roadmap

• Let P = (x, y, z)
• Write expressions for PA² and PB²
• Expand and simplify
• Convert into standard quadratic form
• Identify sphere structure

Solution

Let \(P(x, y, z)\)

\[ PA^2 = (x-3)^2+(y-4)^2+(z-5)^2 \] \[ PB^2 = (x+1)^2+(y-3)^2+(z+7)^2 \]

Adding: \[ PA^2 + PB^2 \]

After expansion: \[ = 2x^2 + 2y^2 + 2z^2 - 4x -14y + 4z + 109 \]

Given: \[ PA^2 + PB^2 = k^2 \]

Divide by 2: \[ x^2 + y^2 + z^2 - 2x - 7y + 2z = \frac{k^2 - 109}{2} \]

This represents a sphere.

Geometric Interpretation

Final Answer

\[ x^2 + y^2 + z^2 - 2x - 7y + 2z = \frac{k^2 - 109}{2} \]

Why This Question Matters

• Classic locus → sphere conversion (very common in CBSE exams)
• Directly used in JEE Main coordinate geometry
• Builds bridge between algebraic form and geometric object
• Tests expansion accuracy + pattern recognition

Exam Insight

Shortcut method: Instead of full expansion, directly use midpoint of A and B as center. This reduces time by ~50% in competitive exams.