1. Chapter at a Glance
Three dimensional geometry extends coordinate geometry to space, where every point has three coordinates (x, y, z). Using vectors, we write equations of lines and planes and find angles and distances between them. It is one of the most reliable scoring chapters of JEE Main — about 2 questions per shift — because questions follow a small set of patterns.
- Shortest distance between two skew lines.
- Foot of the perpendicular and image of a point in a line or a plane.
- Equation of a plane through given points or containing a line.
- Angle between a line and a plane or between two planes; distance of a point from a plane.
2. Coordinates, Direction Cosines and Ratios
- Distance between two points: √((x2 − x1)² + (y2 − y1)² + (z2 − z1)²).
- Section formula works exactly as in 2D, coordinate by coordinate.
- If a line makes angles α, β, γ with the axes, its direction cosines (DCs) are l = cos α, m = cos β, n = cos γ, with l² + m² + n² = 1.
- Direction ratios (DRs) a, b, c are any numbers proportional to l, m, n: l = a/√(a² + b² + c²), and so on.
- DRs of the line through (x1, y1, z1) and (x2, y2, z2): (x2 − x1, y2 − y1, z2 − z1).
- Angle between two lines with DRs (a1, b1, c1) and (a2, b2, c2): cos θ = |a1a2 + b1b2 + c1c2| / (√(a1² + b1² + c1²)√(a2² + b2² + c2²)).
3. Equation of a Line
Here (x1, y1, z1) is a point on the line and (a, b, c) are its direction ratios. Any point on the line is (x1 + aλ, y1 + bλ, z1 + cλ) — this "general point" solves most line problems.
Relation between two lines
- Parallel: direction ratios proportional. Perpendicular: a1a2 + b1b2 + c1c2 = 0.
- Intersecting: shortest distance 0. Skew: neither parallel nor intersecting (possible only in 3D).
For r = a1 + λb1 and r = a2 + μb2:
For parallel lines (b1 = b2 = b): SD = |(a2 − a1) × b| / |b|. The lines are coplanar (intersect) when (a2 − a1) · (b1 × b2) = 0.
4. Equation of a Plane
| Form | Equation |
|---|---|
| General | ax + by + cz + d = 0 (normal has DRs a, b, c) |
| Through a point with normal n | a(x − x1) + b(y − y1) + c(z − z1) = 0 |
| Normal form | lx + my + nz = p (p = distance from origin) |
| Intercept form | x/a + y/b + z/c = 1 |
| Through three points | determinant of (x − x1, y − y1, z − z1; two edge vectors) = 0 |
| Through intersection of two planes | P1 + λP2 = 0 |
Distances and angles involving planes
- Distance of (x1, y1, z1) from ax + by + cz + d = 0: |ax1 + by1 + cz1 + d|/√(a² + b² + c²).
- Distance between parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0: |d1 − d2|/√(a² + b² + c²).
- Angle between planes = angle between their normals.
- Angle φ between a line (DRs b) and a plane (normal n): sin φ = |b · n|/(|b||n|).
- Foot of perpendicular from P to the plane: (x − x1)/a = (y − y1)/b = (z − z1)/c = −(ax1 + by1 + cz1 + d)/(a² + b² + c²); use −2(…) for the image.
5. Formula Master Table
| Result | Formula |
|---|---|
| Direction cosines | l² + m² + n² = 1 |
| Line (Cartesian) | (x − x1)/a = (y − y1)/b = (z − z1)/c |
| Angle between lines | cos θ = |b1·b2|/(|b1||b2|) |
| Skew lines SD | |(a2 − a1)·(b1 × b2)|/|b1 × b2| |
| Parallel lines SD | |(a2 − a1) × b|/|b| |
| Point–plane distance | |ax1 + by1 + cz1 + d|/√(a² + b² + c²) |
| Angle line–plane | sin φ = |b·n|/(|b||n|) |
| Coplanar lines | (a2 − a1)·(b1 × b2) = 0 |
| Plane family | P1 + λP2 = 0 |
6. Mind Map
- DCs: l² + m² + n² = 1
- DRs from two points
- Angle between lines
- r = a + λb
- General point
- Skew lines SD
- Normal (a, b, c)
- Intercept form
- P1 + λP2
- Point–plane distance
- Foot and image
- Line–plane angle (sin)
7. Tips and Tricks
- For the foot of the perpendicular from a point to a line, take the general point Q(λ) on the line and set PQ · b = 0.
- The normal of a plane containing two lines is the cross product of their direction vectors.
- Remember: angle between line and plane uses sin; between two lines or two planes uses cos.
- If a line lies in a plane, its direction is perpendicular to the plane's normal and a point of the line satisfies the plane.
- Forgetting to make the coefficients of x, y, z identical before using the formula for distance between parallel planes.
- Using the plane's normal as the direction of a line lying in the plane (it should be perpendicular to it).
- Mixing up the image (factor 2) and the foot (factor 1).
8. Important One-liners
- The x-axis has direction cosines (1, 0, 0); the line equally inclined to all three axes has DCs (±1/√3, ±1/√3, ±1/√3).
- The equation of the xy-plane is z = 0.
- Two planes are perpendicular if a1a2 + b1b2 + c1c2 = 0.
- Skew lines exist only in three dimensions.
- The distance of the origin from ax + by + cz + d = 0 is |d|/√(a² + b² + c²).
- If a line is parallel to a plane, the distance of any point of the line from the plane is the same.
- cos²α + cos²β + cos²γ = 1 and sin²α + sin²β + sin²γ = 2.
9. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
The direction cosines of a line with direction ratios 2, −1, 2 are:
- (2, −1, 2)
- (2/3, −1/3, 2/3)
- (2/9, −1/9, 2/9)
- (2/√5, −1/√5, 2/√5)
Show solution
√(4 + 1 + 4) = 3, so the DCs are (2/3, −1/3, 2/3).
Answer: (B)
The angle between the planes x + y + z = 1 and x − y = 0 is:
- 0°
- 45°
- 60°
- 90°
Show solution
Normals (1, 1, 1) and (1, −1, 0): dot product = 1 − 1 + 0 = 0, so the normals — and the planes — are perpendicular.
Answer: (D)
Find the distance of the point (1, 2, 3) from the plane 2x − y + 2z = 3.
Show solution
Distance = |2(1) − 2 + 2(3) − 3|/√(4 + 1 + 4) = |3|/3 = 1.
Answer: 1
Find the shortest distance between the lines r = (i + 2j + 3k) + λi and r = μj.
Show solution
a1 = (1, 2, 3), b1 = (1, 0, 0); a2 = (0, 0, 0), b2 = (0, 1, 0).
b1 × b2 = (0, 0, 1) and a2 − a1 = (−1, −2, −3).
SD = |(−1, −2, −3) · (0, 0, 1)|/1 = 3.
Answer: 3
The image of the point (1, 0, 0) in the plane x + y + z = 0 is:
- (−1, 0, 0)
- (1/3, −2/3, −2/3)
- (0, 1, 0)
- (2/3, −1/3, −1/3)
Show solution
(x − 1)/1 = y/1 = z/1 = −2(1 + 0 + 0 + 0)/3 = −2/3.
So x = 1 − 2/3 = 1/3, y = −2/3, z = −2/3. (Option D is the foot of the perpendicular, not the image.)
Answer: (B)
The equation of the plane through (1, 1, 1) perpendicular to the line with direction ratios 2, 3, 4 is:
- 2x + 3y + 4z = 9
- x + y + z = 3
- 2x + 3y + 4z = 0
- 2x − 3y + 4z = 3
Show solution
The line's direction is the plane's normal: 2(x − 1) + 3(y − 1) + 4(z − 1) = 0 ⇒ 2x + 3y + 4z = 9.
Answer: (A)
The foot of the perpendicular from the origin to the plane 2x + 3y + 6z = 49 is (α, β, γ). Find α + β + γ.
Show solution
The foot is t(2, 3, 6) for some t, lying on the plane: 4t + 9t + 36t = 49 ⇒ t = 1.
Foot = (2, 3, 6), so α + β + γ = 11.
Answer: 11
The lines (x − 1)/2 = (y − 2)/3 = (z − 3)/4 and (x − 2)/3 = (y − 4)/4 = (z − 5)/5 are:
- Parallel
- Intersecting
- Skew
- Perpendicular
Show solution
Directions (2, 3, 4) and (3, 4, 5) are not proportional, so the lines are not parallel.
a2 − a1 = (1, 2, 2). b1 × b2 = (3·5 − 4·4, 4·3 − 2·5, 2·4 − 3·3) = (−1, 2, −1).
(1, 2, 2) · (−1, 2, −1) = −1 + 4 − 2 = 1 ≠ 0, so the lines do not intersect: they are skew. (Perpendicular would need 2·3 + 3·4 + 4·5 = 0, which is false.)
Answer: (C)
10. Quick Revision Summary
- l² + m² + n² = 1; DRs are proportional to DCs.
- Line: r = a + λb; use the general point for foot/intersection problems.
- Skew lines SD = |(a2 − a1)·(b1 × b2)|/|b1 × b2|; zero means coplanar.
- Plane normal (a, b, c); point–plane distance |ax1 + by1 + cz1 + d|/√(a² + b² + c²).
- Foot: factor −1; image: factor −2; line–plane angle uses sin.