Three Dimensional Geometry

🟢 Lite — Quick Review

Rapid summary for last-minute revision before your exam.

Three Dimensional Geometry — Key Facts for JEE Main Coordinate system: right-handed system with x, y, z axes Distance between points (x₁, y₁, z₁) and (x₂, y₂, z₂): d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²] Section formula for point dividing P₁P₂ in ratio m:n: (mx₂+nx₁)/(m+n), … Direction ratios: (a, b, c) proportional to direction cosines; direction cosines l = a/|v|, m = b/|v|, n = c/|v| ⚡ Exam tip: JEE Main tests equations of lines and planes, angle between lines/planes — know the standard forms!

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🟡 Standard — Core Study

Standard content for students with a few days to months.

Three Dimensional Geometry — JEE Main Study Guide

Distance formula: d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]

Direction ratios and direction cosines: If vector has direction ratios (a, b, c), direction cosines are: l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²) Note: l² + m² + n² = 1

Equation of line in 3D:

• Vector form: →r = →a + λ→b (through →a in direction →b)
• Cartesian: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c
• Two-point form: (x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁)
• Parametric: x = x₁ + λa, y = y₁ + λb, z = z₁ + λc

Angle between two lines: If direction ratios are (a₁, b₁, c₁) and (a₂, b₂, c₂): cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁²+b₁²+c₁²)·√(a₂²+b₂²+c₂²)] Parallel: a₁/a₂ = b₁/b₂ = c₁/c₂ Perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0

Equation of plane:

• General form: Ax + By + Cz + D = 0 with normal →n = (A, B, C)
• Plane through point (x₁, y₁, z₁): A(x−x₁) + B(y−y₁) + C(z−z₁) = 0
• Plane through three points: use determinant method
• Vector form: →n · (→r − →a) = 0

Angle between two planes: For planes with normals →n₁ = (A₁, B₁, C₁) and →n₂ = (A₂, B₂, C₂): cos θ = |→n₁ · →n₂|/(|→n₁||→n₂|) = |A₁A₂ + B₁B₂ + C₁C₂|/√(…)

Angle between line and plane: For line direction →b = (a, b, c) and plane normal →n = (A, B, C): sin φ = |→n · →b|/(|→n||→b|) where φ is angle between line and normal (so angle between line and plane is 90° − φ)

Distance from point to plane: From point (x₁, y₁, z₁) to plane Ax + By + Cz + D = 0: d = |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²)

Skew lines: Lines that are neither parallel nor intersecting in 3D Shortest distance between skew lines: |(→b₂−→b₁) · (→d₁ × →d₂)| / |→d₁ × →d₂| Where →b₁, →b₂ are position vectors of points on lines, →d₁, →d₂ are direction vectors

Coplanarity: Two lines are coplanar if (→d₁ × →d₂) · (→b₂−→b₁) = 0

• Key formula: Distance between points: √(Δx²+Δy²+Δz²); Line: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c; Plane: Ax+By+Cz+D=0
• Common trap: Angle between line and plane is measured as complement of angle between line and normal
• Exam weight: 1 question per year (4 marks); frequently combined with vector algebra

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🔴 Extended — Deep Dive

Comprehensive coverage for students on a longer study timeline.

Three Dimensional Geometry — Comprehensive JEE Main Notes

Equation of plane in intercept form: If plane cuts x, y, z axes at (a, 0, 0), (0, b, 0), (0, 0, c): x/a + y/b + z/c = 1

Equation of plane through line of intersection: Plane L₁ + λL₂ = 0 passes through intersection of planes L₁ = 0 and L₂ = 0

Image of point in plane: To find image of point P in plane Ax + By + Cz + D = 0: Let foot of perpendicular from P to plane be Q, then image P’ is such that Q is midpoint of PP’ P’ coordinates found using direction ratios proportional to (A, B, C)

Sphere: Equation: (x−h)² + (y−k)² + (z−l)² = r² General form: x² + y² + z² + 2ux + 2vy + 2wz + d = 0 Centre (−u, −v, −w); radius = √(u² + v² + w² − d)

Equation of line perpendicular to plane: If plane is Ax + By + Cz + D = 0, then line through point (x₁, y₁, z₁) perpendicular to plane is: (x−x₁)/A = (y−y₁)/B = (z−z₁)/C

Perpendicular distance between parallel lines: For parallel lines in 3D: d = |→d × (→b₁−→b₂)| / |→d| Where →d is common direction vector, →b₁, →b₂ are points on the two lines

Angle bisector of two lines: Direction ratios of angle bisectors of lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂): (a₁±a₂, b₁±b₂, c₁±c₂) One bisector gives acute angle, one gives obtuse

Angle bisector of two planes: Angle bisectors of planes A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0: (A₁x + B₁y + C₁z + D₁)/√(A₁²+B₁²+C₁²) = ± (A₂x + B₂y + C₂z + D₂)/√(A₂²+B₂²+C₂²)

Centre of sphere from 4 points: Solve system of 4 equations; or use that any 4 non-coplanar points define a unique sphere

Great circle and small circle: Intersection of sphere with plane through centre: great circle All other intersections: small circles

Tangent plane to sphere: At point (x₁, y₁, z₁) on sphere x² + y² + z² = r²: xx₁ + yy₁ + zz₁ = r²

Foot of perpendicular from point to line: For line through →a in direction →d, from point →p: Foot Q = →a + [(→p−→a) · →d/|→d|²] →d Parameter t = (→p−→a) · →d / |→d|²

Intersection of line and plane: Substitute parametric line into plane equation, solve for λ, then get intersection point

Plane through intersection of two planes and parallel to a line: Use plane through line of intersection of planes, then apply condition of being parallel to given line

Family of planes: Plane through point (x₁, y₁, z₁): A(x−x₁) + B(y−y₁) + C(z−z₁) = 0 where (A, B, C) are not all zero This represents all planes through the given point

Condition for line to lie in plane: Line →r = →a + λ→d lies in plane →n · (→r − →b) = 0 if →n · →d = 0 and →n · (→a − →b) = 0

Shortest distance between two skew lines: d = |(→a₂−→a₁) · (→d₁ × →d₂)| / |→d₁ × →d₂| Where →a₁, →a₂ are points on lines, →d₁, →d₂ are direction vectors

Perpendicular from point to plane: If plane is →n · (→r − →a) = 0 and point is →p, then foot of perpendicular: →q = →p − [(→n · (→p−→a)/|→n|²)] →n

• Remember: Line in 3D: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c; Plane: Ax+By+Cz+D=0; Distance point to plane: |Ax₁+By₁+Cz₁+D|/√(A²+B²+C²); Angle line-plane: sin φ = |→n·→d|/(|→n||→d|)
• Previous years: “Find equation of plane through (1,2,3) perpendicular to line (x−1)/2 = (y−2)/3 = (z−3)/4” [2023]; “Find distance between parallel planes 2x + 3y + 4z = 5 and 2x + 3y + 4z = 9” [2024]; “Show that lines (x−1)/2 = (y−2)/3 = (z−3)/4 and (x−1)/1 = (y−2)/1 = (z−3)/1 are skew” [2024]

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📊 JEE Main Exam Essentials

Detail	Value
Questions	90 (30 per subject)
Time	3 hours
Marks	300 (90 per subject)
Section	Physics (30), Chemistry (30), Mathematics (30)
Negative	−1 for wrong answer
Mode	Computer-based

🎯 High-Yield Topics for JEE Main Mathematics

• Calculus (Differentiation + Integration) — ~35 marks combined
• Coordinate Geometry (straight lines, circles, conics) — ~20 marks
• Algebra (Complex Numbers, Quadratics, P&C, Probability) — ~25 marks
• Trigonometry + Inverse Trigonometry — ~15 marks
• Vector + 3D — ~15 marks

📝 Previous Year Question Patterns

• 3D Geometry: 1 question per year, 4 marks
• Common patterns: equation of line/plane, angle between line-plane, distance point to plane, skew lines
• Weight: medium frequency, high scoring

💡 Pro Tips

• For line equations, always identify a point on the line and direction ratios
• For plane equations, the coefficients (A, B, C) in Ax + By + Cz + D = 0 give the normal vector
• To check if two lines intersect, solve the parametric equations; if you get consistent solution, they intersect
• For shortest distance between skew lines, use the triple product formula
• When finding angle between line and plane, use sin φ formula with normal and direction vectors
• Distance between parallel planes: normalise the constant difference by the magnitude of normal vector

🔗 Official Resources

• NTA Official JEE Main
• JEE Main Syllabus PDF

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