JEEClass 12
Vectors & 3D Geometry Formulas for JEE
All Vectors and 3D Geometry formulas for JEE. This is one of the most scoring sections — formulaic problems with predictable patterns. Know these formulas and you're set.
Vector Operations
Dot product⃗a · ⃗b = |a||b|cosθ = a₁b₁ + a₂b₂ + a₃b₃
Cross product magnitude|⃗a × ⃗b| = |a||b|sinθ
Scalar triple product[⃗a ⃗b ⃗c] = ⃗a · (⃗b × ⃗c)
Projectionproj of ⃗a on ⃗b = (⃗a·⃗b)/|⃗b|
Area of triangleA = ½|⃗a × ⃗b|
Area of parallelogramA = |⃗a × ⃗b|
Volume of parallelepipedV = |[⃗a ⃗b ⃗c]|
Coplanarity test[⃗a ⃗b ⃗c] = 0
3D Line Equations
Vector form⃗r = ⃗a + λ⃗b
Cartesian form(x−a₁)/b₁ = (y−a₂)/b₂ = (z−a₃)/b₃
Through two points⃗r = ⃗a + λ(⃗b − ⃗a)
3D Plane Equations
Normal form⃗r · ⃗n = d
Cartesian formax + by + cz = d
Intercept formx/a + y/b + z/c = 1
Through 3 pointsUse determinant of (r−a, b−a, c−a) = 0
Distances & Angles
Point to planed = |⃗r₁ · ⃗n − d| / |⃗n|
Between parallel planesd = |d₁ − d₂| / |⃗n|
Skew lines shortest dist.d = |(⃗a₂−⃗a₁) · (⃗b₁×⃗b₂)| / |⃗b₁×⃗b₂|
Angle: line & planesinθ = |⃗b · ⃗n| / (|⃗b||⃗n|)
Angle: two planescosθ = |⃗n₁ · ⃗n₂| / (|⃗n₁||⃗n₂|)
Image of point in plane(x−x₁)/a = (y−y₁)/b = (z−z₁)/c = −2(ax₁+by₁+cz₁−d)/(a²+b²+c²)
JEE Tips
- Tip 1:Shortest distance between skew lines is the most common JEE question from this chapter
- Tip 2:Remember: dot product gives scalar (angles), cross product gives vector (areas)
- Tip 3:For image of point in plane: find foot of perpendicular first, then use midpoint formula