🎯 Key Points
- Direction cosines satisfy l²+m²+n²=1 always (they're cosines of angles a UNIT vector makes with the axes); direction ratios are just any proportional set, not normalized
- Lines: perpendicular when b₁·b₂=0 (dot product); parallel when b₁×b₂=0 (cross product) — same pattern as the dot/cross distinction in Vector Algebra
- Shortest distance between skew lines = |(c−a)·(b×d)|/|b×d| — this formula only applies to skew (non-intersecting, non-parallel) lines; intersecting lines have distance 0
- Point-to-plane distance |ax₀+by₀+cz₀+d|/√(a²+b²+c²) directly parallels the 2D point-to-line distance formula, just with one extra coordinate
Three Dimensional Geometry
Extends coordinate geometry to 3D space. Essential for understanding physical space, engineering, and computer graphics.
Direction Cosines and Ratios
Three mutually perpendicular axes x, y, z meeting at the origin O, with a point P located by its (x, y, z) coordinates.
- Direction cosines l, m, n: cosines of angles with x, y, z axes
- l² + m² + n² = 1 (fundamental identity)
- Direction ratios a, b, c are proportional to l, m, n
- l = a/√(a²+b²+c²) etc.
Line in 3D
- Vector form: r = a + λb (a: fixed point, b: direction vector)
- Cartesian form: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c
- Angle between lines: cos θ = |b₁·b₂| / (|b₁||b₂|)
- Perpendicular: b₁·b₂ = 0 | Parallel: b₁ × b₂ = 0
Skew Lines
Non-coplanar, non-intersecting, non-parallel lines in 3D.
Shortest distance = |(c − a) · (b × d)| / |b × d|
Plane
- General form: ax + by + cz + d = 0; normal = (a,b,c)
- Intercept form: x/a + y/b + z/c = 1
- Normal form: lx + my + nz = p
Distances and Angles
- Point to plane: |ax₀+by₀+cz₀+d| / √(a²+b²+c²)
- Angle between planes: cos θ = |n₁·n₂| / (|n₁||n₂|)
- Angle between line and plane: sin θ = |b·n| / (|b||n|)
- Coplanarity of two lines: (c−a)·(b×d) = 0
Coordinates and Distance Formula in 3D
A point in space is fixed by an ordered triple (x, y, z) relative to three mutually perpendicular axes. The three coordinate planes (xy, yz, zx) divide space into eight octants.
- Distance between two points P(x₁,y₁,z₁) and Q(x₂,y₂,z₂): PQ = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²].
- Distance from the origin: OP = √(x² + y² + z²).
- Example: distance between (1, −2, 3) and (4, 2, 3) = √[9 + 16 + 0] = 5.
Section Formula in 3D
The point R dividing the segment joining P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) in the ratio m : n has coordinates:
- Internal division: R = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n), (mz₂ + nz₁)/(m+n)).
- External division: replace n by −n, giving ((mx₂ − nx₁)/(m−n), …).
- Midpoint (ratio 1:1): ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2).
- Centroid of a triangle with vertices A, B, C: ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3).
Equation of a Plane — Point-Normal and Three-Point Forms
- Point-normal (vector) form: a plane through point A with normal vector n is n·(r − a) = 0. If n = (A, B, C) and the point is (x₁,y₁,z₁), this expands to A(x−x₁) + B(y−y₁) + C(z−z₁) = 0.
- Plane through three points A, B, C: take n = AB × AC as the normal, then apply the point-normal form. Equivalently the coplanarity determinant of (r−A), (B−A), (C−A) equals 0.
- Distance of the plane from the origin for ax+by+cz+d = 0 is |d|/√(a²+b²+c²).
Plane Through the Intersection of Two Planes
If two planes are P₁: a₁x+b₁y+c₁z+d₁ = 0 and P₂: a₂x+b₂y+c₂z+d₂ = 0, then for any real λ the equation
(a₁x+b₁y+c₁z+d₁) + λ(a₂x+b₂y+c₂z+d₂) = 0
represents a plane through their line of intersection. Choosing λ to satisfy one extra condition (passing through a given point, being perpendicular to another plane, etc.) pins down the required plane without first finding the line of intersection explicitly.
🚀 JEE Advanced Edge
Finding the equation of a plane through three points without memorizing a separate formula: Given three points A, B, C, the plane through them can be found by computing two direction vectors AB and AC, taking their cross product to get the normal vector n=AB×AC, then using the point-normal form n·(r−A)=0 — this reduces "plane through 3 points" to vector operations you already know, rather than a determinant formula to memorize separately.
Why the angle between a line and a plane uses sine, not cosine: The angle between two LINES (or two planes, via their normals) is measured between the direction vectors directly, using cosine — but the angle between a line and a plane is measured between the line and the plane's SURFACE, which is the complement of the angle between the line and the plane's normal, converting the cosine formula into a sine formula: sinθ=|b·n|/(|b||n|).
Worked problem: Find the foot of the perpendicular from the point (1,2,3) to the plane x+y+z=6, and use it to find the distance from the point to the plane. Approach: The plane's normal is (1,1,1). Distance = |1+2+3−6|/√(1²+1²+1²) = |0|/√3 = 0. The point (1,2,3) already lies exactly on the plane (1+2+3=6), so the foot of the perpendicular is the point itself and the distance is 0.
Worked Example: Distance from a Point to a Plane
Find the distance from the point (2, 3, −1) to the plane 2x − y + 2z = 4.
Distance formula: d = |ax₁ + by₁ + cz₁ − d| / √(a² + b² + c²) = |2(2) − (3) + 2(−1) − 4| / √(4+1+4) = |4 − 3 − 2 − 4| / 3 = |−5| / 3 = 5/3 units. The formula always gives a non-negative result — include the absolute value.
Worked Example: Equation of a Line Through Two Points
Find the vector equation of the line passing through A(1, 2, 3) and B(4, 6, 5), and determine a point on it at parameter t = 2.
Direction vector: b = B − A = 3î + 4ĵ + 2k̂. Vector equation: r = (î + 2ĵ + 3k̂) + t(3î + 4ĵ + 2k̂). At t = 2: r = (1+6)î + (2+8)ĵ + (3+4)k̂ = (7, 10, 7). In Cartesian form: (x−1)/3 = (y−2)/4 = (z−3)/2.