Three Dimensional Geometry extends coordinate geometry into space. Every point now needs three coordinates (x, y, z), and lines and planes are described by equations in all three. For NDA, the most tested results are: distance formula in 3D, direction cosines and ratios, the symmetric form of a line, the equation of a plane, and the perpendicular distance from a point to a plane. Conceptually it is the same as 2D but with one extra dimension.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Distance between two points in 3D using the extended distance formula; (2) Finding direction cosines l, m, n and verifying l²+m²+n²=1; (3) Converting direction ratios (a, b, c) to direction cosines; (4) Equation of a line through a given point with given direction ratios (symmetric form); (5) Equation of a plane in general form Ax+By+Cz+D=0 and intercept form; (6) Angle between two lines using their direction cosines or ratios; (7) Condition for two lines to be parallel or perpendicular; (8) Perpendicular distance from a point to a plane (most directly tested formula).
Topics at a Glance
① 3D Coordinates & Distance
Distance, midpoint, section formula in 3D
② Direction Cosines & Ratios
l,m,n; l²+m²+n²=1; ratios a,b,c
③ Equation of a Line
Symmetric, two-point, vector form
④ Equation of a Plane
General, intercept, normal form
⑤ Angles & Relations
Angle between lines, lines and planes
⑥ Point–Plane Distance
|Ax₀+By₀+Cz₀+D|/√(A²+B²+C²)
1. 3D Coordinates — Distance, Midpoint & Section Formula
1.1
Extending 2D Formulas into Three Dimensions
Every 2D formula gains a third term — (z₂−z₁)² in distance, z-term in section formula
⚡ 3D Coordinate Basics
DISTANCE between P(x₁,y₁,z₁) and Q(x₂,y₂,z₂):
PQ = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]
MIDPOINT of PQ:
M = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)
SECTION FORMULA — point R dividing PQ in ratio m:n internally:
R = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n), (mz₂+nz₁)/(m+n))
CENTROID of tetrahedron with vertices (xᵢ,yᵢ,zᵢ), i=1..4:
G = ((x₁+x₂+x₃+x₄)/4, (y₁+y₂+y₃+y₄)/4, (z₁+z₂+z₃+z₄)/4)
Distance from origin to P(x,y,z):
OP = √(x²+y²+z²)
The 3D distance formula is a direct extension of 2D: add the square of the z-difference to the sum. Everything else follows the same pattern — just add the third coordinate wherever needed.
Fig 1: 3D coordinate system with x-axis (green), y-axis (amber), z-axis (red). Point P(x,y,z) with dashed projection lines to the axes. Every 2D distance/midpoint formula gains a third z-term.
Worked Example — Distance in 3D
Find the distance between A(1,2,3) and B(4,6,3).
AB = √[(4−1)²+(6−2)²+(3−3)²] = √[9+16+0] = √25 = 5.
📝 TOPIC-WISE PYQ
3D Basics — NDA-Pattern Questions
Q1. Distance from origin to point (3, 4, 0) is:
(a) 3 (b) 4 (c) 5 (d) 7
Answer: (c) 5
d=√(9+16+0)=√25=5.
Q2. The midpoint of the segment joining (2,4,6) and (0,−2,−4) is:
Direction Cosines l, m, n — The Angles a Line Makes with Axes
l²+m²+n²=1 is the fundamental identity; direction ratios are proportional to DCs
⚡ Direction Cosines & Ratios — Complete Set
DIRECTION COSINES (l, m, n):
l = cosα, m = cosβ, n = cosγ
where α, β, γ are the angles made with x, y, z axes respectively.
FUNDAMENTAL IDENTITY: l² + m² + n² = 1 (always!)
DIRECTION RATIOS (a, b, c):
Any three numbers proportional to l, m, n are called direction ratios.
If DRs are (a, b, c), then direction cosines are:
l = a / √(a²+b²+c²)
m = b / √(a²+b²+c²)
n = c / √(a²+b²+c²)
The denominator √(a²+b²+c²) normalises the direction ratios.
DIRECTION COSINES OF A LINE THROUGH TWO POINTS P(x₁,y₁,z₁) and Q(x₂,y₂,z₂):
DRs: (x₂−x₁, y₂−y₁, z₂−z₁) = (Δx, Δy, Δz)
DCs: l=(Δx)/d, m=(Δy)/d, n=(Δz)/d where d = PQ
SPECIAL DIRECTION COSINES (axes):
x-axis: (1,0,0) y-axis: (0,1,0) z-axis: (0,0,1)
Line making equal angles with all axes: l=m=n=1/√3
Direction cosines are UNIQUE (given a fixed direction). Direction ratios are NOT unique — any scalar multiple gives the same direction. The normalisation (dividing by √(a²+b²+c²)) converts ratios to cosines.
Key Identities with DCs
l²+m²+n²=1 — always, no exception
If l=m=n: 3l²=1 → l=±1/√3
cos²α+cos²β+cos²γ=1
sin²α+sin²β+sin²γ=2 (derived)
DCs of x-axis: α=0°,β=90°,γ=90° → (1,0,0)
Converting DRs to DCs
DRs (2,3,6) → norm=√(4+9+36)=7
DCs = (2/7, 3/7, 6/7)
Check: 4/49+9/49+36/49=49/49=1 ✓
Negative DRs give reverse direction DCs
Parallel lines have same (or ±) DCs
Worked Example — Direction Cosines
Find the DCs of the line through A(1,2,3) and B(5,6,9).
🤯 T1. A line makes angles α, β, γ with axes. Show sin²α+sin²β+sin²γ=2.
l²+m²+n²=1 → cos²α+cos²β+cos²γ=1.
sin²α=1−cos²α; similarly for β, γ.
Sum = (1−cos²α)+(1−cos²β)+(1−cos²γ) = 3−(cos²α+cos²β+cos²γ) = 3−1 = 2 ✓.
🤯 T2. If a line makes equal angles with all three coordinate axes, find those angles.
Let each angle be α. Then l=m=n=cosα.
l²+m²+n²=1 → 3cos²α=1 → cosα=±1/√3.
α = cos⁻¹(1/√3) ≈ 54.74°. There are two such lines (one for each sign), giving angles ≈54.74° and 125.26° with each axis.
3. Equation of a Line in 3D
3.1
Symmetric & Vector Forms of a Line
Symmetric form is most common in NDA; vector form is elegant for theoretical work
⚡ Equation of a Line in 3D — All Forms
SYMMETRIC FORM (through point (x₁,y₁,z₁), direction ratios a,b,c):
(x−x₁)/a = (y−y₁)/b = (z−z₁)/c = λ (parameter)
Each part equals λ: x = x₁+aλ, y = y₁+bλ, z = z₁+cλ
TWO-POINT FORM (through P(x₁,y₁,z₁) and Q(x₂,y₂,z₂)):
(x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁)
VECTOR FORM:
If point A has position vector a⃗, and direction vector is b⃗:
r⃗ = a⃗ + λb⃗ (λ is a scalar parameter, −∞<λ<∞)
SPECIAL LINES:
Line along x-axis: y=0, z=0 ⇒ x/1=y/0=z/0 (DRs 1,0,0)
Line parallel to y-axis through (x₀,y₀,z₀): (x−x₀)/0=(y−y₀)/1=(z−z₀)/0
NOTES:
If one denominator is 0 (say a=0): x=x₁ (a constraint, not division by zero)
The three ratios each equal the same parameter λ
The symmetric form is the standard form for NDA. When one of a, b, c is zero, the corresponding variable is fixed (e.g., a=0 means x=x₁ throughout). Always identify the point on the line and its direction ratios.
Worked Example — Line in Symmetric Form
Write the equation of the line through (2,−1,3) with DRs (3,4,−2).
(x−2)/3 = (y+1)/4 = (z−3)/(−2).
This is the symmetric form. Any point on the line: x=2+3λ, y=−1+4λ, z=3−2λ.
Worked Example — Two-Point Form
Line through A(1,2,3) and B(4,5,7).
DRs = (4−1,5−2,7−3) = (3,3,4).
(x−1)/3 = (y−2)/3 = (z−3)/4.
📝 TOPIC-WISE PYQ
Equation of a Line — NDA-Pattern Questions
Q6. The direction ratios of the line (x−1)/2 = (y+2)/3 = (z−3)/4 are:
(a) 1,2,3 (b) 2,3,4 (c) −1,2,−3 (d) 1,−2,3
Answer: (b) 2,3,4
The denominators in the symmetric form are the DRs: (2,3,4).
Q7. The equation of a line through origin with DRs (2,3,6) is:
Answer: (a) x/2=y/3=z/6
Through (0,0,0) with DRs (2,3,6): (x−0)/2=(y−0)/3=(z−0)/6 → x/2=y/3=z/6.
4. Equation of a Plane
4.1
General, Intercept & Normal Forms
The normal vector (A,B,C) in Ax+By+Cz+D=0 is perpendicular to every line in the plane
⚡ Equations of a Plane — All Forms
GENERAL FORM:
Ax + By + Cz + D = 0
(A,B,C) = normal vector to the plane
If D=0: plane passes through origin
PLANE THROUGH POINT (x₀,y₀,z₀) with normal (A,B,C):
A(x−x₀) + B(y−y₀) + C(z−z₀) = 0
INTERCEPT FORM (cuts x-axis at a, y at b, z at c):
x/a + y/b + z/c = 1
NORMAL FORM (l,m,n are DCs of normal, p=distance from origin):
lx + my + nz = p (p ≥ 0)
From general: divide Ax+By+Cz+D=0 by −D/√(A²+B²+C²)
PLANE THROUGH THREE POINTS:
Use determinant: |x−x₁ y−y₁ z−z₁| = 0
|x₂−x₁ y₂−y₁ z₂−z₁|
|x₃−x₁ y₃−y₁ z₃−z₁|
SPECIAL PLANES:
xy-plane (z=0): A=0,B=0,C=1,D=0
yz-plane (x=0): A=1,B=0,C=0,D=0
xz-plane (y=0): A=0,B=1,C=0,D=0
The normal to the plane Ax+By+Cz+D=0 is the vector (A,B,C). This vector is perpendicular to the plane. Two planes are parallel if their normals are parallel ((A₁,B₁,C₁) ∝ (A₂,B₂,C₂)) and perpendicular if the dot product of normals equals zero.
Fig 2: A plane in 3D space. x-intercept a (green), y-intercept b (amber), z-intercept c (red). The purple arrow is the normal vector (A,B,C) — perpendicular to the entire plane.
Worked Example — Intercept Form to General
Plane cuts x,y,z axes at 2,3,6 respectively. Find general form.
x/2+y/3+z/6=1. Multiply by 6: 3x+2y+z=6. General: 3x+2y+z−6=0.
Worked Example — Plane Through a Point with Normal
Plane through (1,2,3) with normal (2,−1,4).
2(x−1)−1(y−2)+4(z−3)=0 → 2x−2−y+2+4z−12=0.
General: 2x−y+4z−12=0.
📝 TOPIC-WISE PYQ
Equation of a Plane — NDA-Pattern Questions
Q8. The plane x/3+y/4+z/6=1 makes intercepts on axes. The y-intercept is:
(a) 3 (b) 4 (c) 6 (d) 12
Answer: (b) 4
In x/a+y/b+z/c=1, the y-intercept is b=4.
Q9. The normal to the plane 3x−2y+5z−7=0 is in the direction:
Answer: (b) (3,−2,5)
From Ax+By+Cz+D=0, normal = (A,B,C) = (3,−2,5).
5. Angles Between Lines and Planes, & Point–Plane Distance
5.1
Angle Formulas & Perpendicular Distance
All involve dot products — angle between lines uses DCs, distance uses normal magnitude
⚡ Angles & Distance — Complete Formula Set
ANGLE BETWEEN TWO LINES (DCs l₁,m₁,n₁ and l₂,m₂,n₂):
cosθ = |l₁l₂ + m₁m₂ + n₁n₂| (acute angle version)
PARALLEL lines: l₁/l₂ = m₁/m₂ = n₁/n₂ (DRs proportional)
PERPENDICULAR: l₁l₂ + m₁m₂ + n₁n₂ = 0 (dot product = 0)
Using DRs (a₁,b₁,c₁) and (a₂,b₂,c₂):
cosθ = |a₁a₂+b₁b₂+c₁c₂| / [√(a₁²+b₁²+c₁²) ⋅ √(a₂²+b₂²+c₂²)]
Perp condition: a₁a₂+b₁b₂+c₁c₂ = 0
ANGLE BETWEEN TWO PLANES Ax+By+Cz+D=0 and A′x+B′y+C′z+D′=0:
cosθ = |AA′+BB′+CC′| / [√(A²+B²+C²) ⋅ √(A′²+B′²+C′²)]
PARALLEL planes: A/A′ = B/B′ = C/C′ (normals parallel)
PERPENDICULAR planes: AA′+BB′+CC′ = 0 (normals perpendicular)
DISTANCE from point P(x₀,y₀,z₀) to plane Ax+By+Cz+D=0:
d = |Ax₀ + By₀ + Cz₀ + D| / √(A²+B²+C²)
DISTANCE from ORIGIN to plane:
d = |D| / √(A²+B²+C²)
DISTANCE BETWEEN TWO PARALLEL PLANES Ax+By+Cz+D₁=0 and D₂=0:
d = |D₁−D₂| / √(A²+B²+C²)
All these formulas use the same geometric idea: angle between directions (dot product / magnitudes product) and distance (numerator has the point substituted, denominator is the magnitude of the normal vector). Master these three formulas and you can answer every NDA relations question in this chapter.
Worked Example — Angle Between Lines
Find angle between lines with DRs (1,2,2) and (2,2,1).
Find distance from (2,3,−1) to plane 3x+4y+5z−10=0.
d = |3(2)+4(3)+5(−1)−10|/√(9+16+25) = |6+12−5−10|/√50 = 3/(5√2) = 3√2/10.
📝 TOPIC-WISE PYQ
Angles & Distance — NDA-Pattern Questions
Q10. Two lines have DRs (1,−1,0) and (0,1,−1). The angle between them is:
(a) 30° (b) 45° (c) 60° (d) 90°
Answer: (c) 60°
cosθ=|0−1+0|/(√2⋅√2)=1/2. θ=60°.
Q11. Distance from origin to plane 2x+3y+6z−21=0 is:
(a) 21/7 (b) 3 (c) 7 (d) 21
Answer: (b) 3
d=|−21|/√(4+9+36)=21/7=3.
Q12. Two planes 2x+y−z=5 and x−2y+3z=1 are:
(a) Parallel (b) Perpendicular (c) Coincident (d) None of these
Answer: (b) Perpendicular
AA′+BB′+CC′=2(1)+1(−2)+(−1)(3)=2−2−3=−3 ≠0. Hmm, let me recalculate:
A₁=2,B₁=1,C₁=−1; A₂=1,B₂=−2,C₂=3.
Dot: 2(1)+1(−2)+(−1)(3)=2−2−3=−3 ≠ 0. Not perpendicular.
Ratio: 2/1≠1/−2 not parallel. Answer: (d) None of these. They intersect at an angle.
Q13. Distance from point (1,2,3) to plane x+2y+2z=3 is:
🤯 T3. Find the equation of the plane through (1,2,1), (2,3,4) and (3,1,2).
Use the three-point determinant form. Let (x₁,y₁,z₁)=(1,2,1).
DRs of vectors in plane: (2−1,3−2,4−1)=(1,1,3) and (3−1,1−2,2−1)=(2,−1,1).
Normal = (1,1,3)×(2,−1,1): n⃗=(1⋅1−3⋅(−1), 3⋅2−1⋅1, 1⋅(−1)−1⋅2)=(1+3,6−1,−1−2)=(4,5,−3).
Plane: 4(x−1)+5(y−2)−3(z−1)=0 → 4x+5y−3z−4−10+3=0. 4x+5y−3z−11=0. Always find the normal by cross product of two vectors lying in the plane.
🤯 T4. Show that the lines (x−1)/2=(y−2)/3=(z−3)/4 and (x−4)/5=(y−1)/2=(z)/(1) are not parallel and find the angle between them.
DRs of L₁: (2,3,4). DRs of L₂: (5,2,1).
For parallel: 2/5≠3/2 → NOT parallel.
cosθ=|2(5)+3(2)+4(1)|/[√(4+9+16)√(25+4+1)]
=|10+6+4|/(√29√30)=20/√870.
θ=cos⁻¹(20/√870)≈cos⁻¹(0.678)≈47.3°. Always check parallel condition first (ratios of DRs), then compute dot product for angle.
🤯 T5. Find the distance between the parallel planes 2x−y+3z+4=0 and 4x−2y+6z−10=0.
Verify parallel: divide second by 2: 2x−y+3z−5=0. Same A,B,C as first (2,−1,3).
d=|D₁−D₂|/√(A²+B²+C²).
D₁=4, D₂ (from 2x−y+3z−5=0) = −5.
d=|4−(−5)|/√(4+1+9)=9/√14=9/√14 = 9√14/14. Key: always divide parallel planes to the same A,B,C coefficient scale before applying the formula.
📝 Master Formula Sheet — MN23 Three Dimensional Geometry
All critical formulae for rapid pre-exam revision.
◆ 3D Basics
Distance=√[(Δx)²+(Δy)²+(Δz)²]
Midpoint=((x₁+x₂)/2,(y₁+y₂)/2,(z₁+z₂)/2)
Section(m:n): (mx₂+nx₁)/(m+n) each coord
Distance from O: √(x²+y²+z²)
⎯ Direction Cosines/Ratios
l=cosα, m=cosβ, n=cosγ
l²+m²+n²=1 (always!)
DRs(a,b,c): l=a/√(a²+b²+c²)
sin²α+sin²β+sin²γ=2
Equal angles with axes: l=m=n=1/√3
◑ Equation of Line
Symmetric: (x−x₁)/a=(y−y₁)/b=(z−z₁)/c
Two-point: use DRs=(x₂−x₁,y₂−y₁,z₂−z₁)
Vector: r⃗=a⃗+λb⃗
If a=0: x=x₁ (fixed), not division by zero
▲ Equation of Plane
General: Ax+By+Cz+D=0
Normal=(A,B,C)
Intercept: x/a+y/b+z/c=1
Through point: A(x−x₀)+B(y−y₀)+C(z−z₀)=0
∠ Angle Formulas
Between lines: cosθ=|l₁l₂+m₁m₂+n₁n₂|
Perp lines: a₁a₂+b₁b₂+c₁c₂=0
Between planes: cosθ=|AA′+BB′+CC′|/(norm₁⋅norm₂)
Perp planes: AA′+BB′+CC′=0
⊥ Distance Formulas
Point to plane: |Ax₀+By₀+Cz₀+D|/√(A²+B²+C²)
Origin to plane: |D|/√(A²+B²+C²)
Between parallel planes: |D₁−D₂|/√(A²+B²+C²)
Parallel: same A,B,C ratio; perp: dot=0
⚡ Quick Revision Booster — MN23 Three Dimensional Geometry
◆ 3D vs 2D
Distance: add (z₂−z₁)² inside root
Midpoint: average all three separately
Section: same ratio formula for z too
Everything else is direct extension
⎯ Direction Cosines
l²+m²+n²=1 (mandatory check)
DRs→DCs: divide each by √(a²+b²+c²)
Equal angles: l=m=n=1/√3
sin²α+sin²β+sin²γ=2
◑ Line & Plane Setup
Line: identify point + DRs → write symmetric form
Plane: identify normal (A,B,C) + a point
Normal of Ax+By+Cz+D=0 is (A,B,C)
Intercept form: read off a,b,c directly
∠ Parallel & Perpendicular
Parallel lines: DRs proportional
Perp lines: dot product of DRs = 0
Parallel planes: normals proportional
Perp planes: normals dot = 0
⊥ Distance Formula
Substitute (x₀,y₀,z₀) into Ax+By+Cz+D
Take absolute value of result
Divide by √(A²+B²+C²)
For origin: numerator=|D| only
🚨 Critical Exam Traps
l²+m²+n²=1 NOT equal to a²+b²+c²
Normal of Ax+By+Cz+D=0 is (A,B,C) not (−A,−B,−C)
Zero denominator in symmetric form means that variable is fixed
Parallel planes: scale to same A,B,C before distance formula
Angle between planes = angle between their normals
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