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Three-dimensional geometry has a reputation for messy formulas: direction cosines, symmetric forms, shortest distances. In practice almost every question reduces to two tools you already know from vectors, the dot product and the cross product. This post shows how each formula comes from those two tools, works through the main question types, and points out which topics belong to which exam.
What is in the JEE Main and JEE Advanced syllabus?
The JEE (Main) 2026 syllabus lists coordinates of a point in space, distance, section formula, direction ratios and direction cosines, the angle between two intersecting lines, the equation of a line, and "Skew lines, the shortest distance between them and its equation". It does not list planes. The JEE (Advanced) 2026 brochure covers lines and adds the equation of a plane, the distance of a point from a plane, angles between two planes and between a line and a plane, and coplanar lines.
NCERT's Class 12 chapter on three dimensional geometry works through direction cosines, lines, the angle between lines and the shortest distance between them. If you are preparing for JEE Advanced, add planes from a reference book. Check the syllabus for your own exam year, since these lists change.
How do direction ratios and direction cosines differ?
Direction ratios (a, b, c) are any numbers proportional to the direction. Direction cosines (l, m, n) are the components of the unit vector, so l² + m² + n² = 1. To convert, divide by the length.
Example. Direction ratios (2, −1, 2) have length √(4 + 1 + 4) = 3, so the direction cosines are (2/3, −1/3, 2/3), or the negatives of all three for the opposite direction.
The angle between two lines comes from the dot product of their directions: cos θ = |b₁·b₂| / (|b₁||b₂|). For b₁ = (1, 2, 2) and b₂ = (2, −2, 1), the dot product is 2 − 4 + 2 = 0, so the lines are perpendicular.
How do you find the foot of a perpendicular?
Take the line through A(1, 0, 2) with direction b = (1, 2, 2), and the point P(2, 0, 6).
A general point on the line is F = (1 + t, 2t, 2 + 2t).
PF = (t − 1, 2t, 2t − 4) must be perpendicular to b, so (t − 1) + 4t + (4t − 8) = 0, giving 9t = 9 and t = 1.
The foot is F = (2, 2, 4), and the distance PF = |(0, 2, −2)| = √8 = 2√2.
Check: (0, 2, −2)·(1, 2, 2) = 0 + 4 − 4 = 0. The same three steps give the image of a point in a line: go twice as far, to 2F − P.
How do you find the shortest distance between skew lines?
Skew lines are neither parallel nor intersecting. The shortest segment between them is perpendicular to both, so it points along b₁ × b₂. Project the vector joining any two points, one on each line, onto that direction:
d = |(a₂ − a₁)·(b₁ × b₂)| / |b₁ × b₂|
Worked example. Line 1 passes through (1, 1, 0) with direction (2, −1, 1). Line 2 passes through (2, 1, −1) with direction (3, −5, 2).
b₁ × b₂ = ((−1)(2) − (1)(−5), (1)(3) − (2)(2), (2)(−5) − (−1)(3)) = (3, −1, −7).
|b₁ × b₂| = √(9 + 1 + 49) = √59.
a₂ − a₁ = (1, 0, −1), and (1, 0, −1)·(3, −1, −7) = 3 + 0 + 7 = 10.
d = 10/√59, about 1.30 units.
If the answer is zero, the lines intersect. If b₁ × b₂ is the zero vector, the lines are parallel and the formula breaks: use d = |b × (a₂ − a₁)| / |b| instead, which NCERT also derives.
How do planes work? (JEE Advanced)
A plane with normal n = (a, b, c) has equation ax + by + cz + d = 0. The distance from a point (x₀, y₀, z₀) is the projection of a joining vector onto n:
distance = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²)
Example. The distance from (1, 2, 3) to 2x − y + 2z + 3 = 0 is |2 − 2 + 6 + 3| / √(4 + 1 + 4) = 9/3 = 3.
For the angle φ between a line and a plane, use the sine: sin φ = |b·n| / (|b||n|), because the normal is perpendicular to the plane. For b = (1, 2, 2) and n = (2, −1, 2), b·n = 2 − 2 + 4 = 4 and |b||n| = 9, so sin φ = 4/9.
Which mistakes cost the most marks?
Using direction ratios as direction cosines without dividing by the length.
Using cosine for a line-plane angle. The line-plane angle uses sine; the angle between two planes uses the cosine of the angle between their normals.
Applying the skew-line formula to parallel lines, which divides by zero.
Dropping the absolute value and getting a negative distance.
Cross product sign errors in the middle component. Write the determinant out every time.
The vector habits in our post on common vector errors apply directly here. If you are still deciding how much JEE Advanced material to cover, read how JEE Main and JEE Advanced differ.
A practice routine
Week 1: direction cosines, angles between lines and foot of the perpendicular. Ten problems, each checked with a dot product.
Week 2: skew and parallel lines. For each pair, first decide which case it is, then apply the formula.
Week 3 (for JEE Advanced): planes, including distance, angles and the line of intersection of two planes.
Week 4: a mixed set of fifteen from all three weeks, timed, plus five from vector algebra.
If you can follow the algebra but cannot picture the lines, Fermi's Simulation Lab has interactive simulations for ideas that are easier to see than to read.
Today, change the direction of line 2 in the worked example to (4, −2, 2) and decide which formula you now need before calculating anything.
