Three-Dimensional Geometry

Three-dimensional geometry is one of the highest-weighted topics in JEE Mathematics, consistently yielding 3-4 questions. It deals with lines, planes, and their relationships in 3D space using both vector and Cartesian forms.

A straight line in 3D is defined by a point and a direction. The vector equation of a line through point a with direction b is r = a + $\lambda$·b. The Cartesian form is (x-x₁)/a = (y-y₁)/b = (z-z₁)/c, where (a, b, c) are direction ratios. A line through two points a and b is r = a + $\lambda$(b - a).

A plane is defined by a point and a normal vector. The vector equation is r . n = d, where n is the normal. The Cartesian form is ax + by + cz = d, where (a, b, c) are the components of n. A plane through three non-collinear points a, b, c satisfies (r - a) . [(b - a) x (c - a)] = 0. The intercept form is x/a + y/b + z/c = 1.

The angle between two lines with directions b₁ and b₂ is cos($\theta$) = |b₁.b₂|/(|b₁||b₂|). Lines are perpendicular when b₁.b₂ = 0 and parallel when b₁ x b₂ = 0. The angle between two planes with normals n₁ and n₂ is cos($\theta$) = |n₁.n₂|/(|n₁||n₂|). The angle between a line and a plane is sin($\phi$) = |b.n|/(|b||n|).

The distance from a point to a plane r.n = d is |a.n - d|/|n|. The distance from a point to a line through a with direction b is |(a - p) x b|/|b|, where p is the point.

The shortest distance between two skew lines r = a₁ + $\lambda$b₁ and r = a₂ + $\mu$b₂ is |(a₂ - a₁) . (b₁ x b₂)|/|b₁ x b₂|. If this distance is zero, the lines are coplanar (intersecting or parallel).

The image of a point P in a plane ax+by+cz=d is found by: (1) draw the perpendicular from P to the plane, (2) find the foot, (3) extend by equal distance. Formula: (x-x₁)/a = (y-y₁)/b = (z-z₁)/c = -2(ax₁+by₁+cz₁-d)/(a²+b²+c²).

The key problem-solving concept is choosing the right form (vector vs Cartesian) based on the problem, and applying distance/angle formulas systematically while keeping track of absolute values for distances and angles.

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