Chapter 11 — Three Dimensional Geometry

Overview: In three-dimensional (3D) space each point is represented by an ordered triple (x, y, z) giving its distances from three mutually perpendicular coordinate axes: x-axis, y-axis and z-axis. The origin O is (0,0,0). The coordinate system divides space into eight regions called octants.

Position vector: The position vector of a point P(x, y, z) is the vector from the origin O to P. It is written as r = OP = xi + yj + zk, where i, j, k are unit vectors along the x-, y-, z-axes respectively. The position vector uniquely identifies the location of the point in vector form.

Vector between two points: If P(x1, y1, z1) and Q(x2, y2, z2) then the vector from P to Q is
PQ = Q - P = (x2 - x1)i + (y2 - y1)j + (z2 - z1)k.

Length (magnitude): The magnitude of the position vector r = xi + yj + zk equals the distance of P from the origin:
|r| = sqrt(x^2 + y^2 + z^2).

Unit vector: The unit vector in the direction of r is
u = r / |r| = (x / |r|) i + (y / |r|) j + (z / |r|) k.

Direction cosines: If α, β, γ are the angles that vector r makes with the x-, y-, z-axes then
cos α = x / |r|, cos β = y / |r|, cos γ = z / |r|,
and cos^2α + cos^2β + cos^2γ = 1.

Midpoint and distance between two points:
Midpoint of P(x1,y1,z1) and Q(x2,y2,z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).
Distance PQ = sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2).

Projections on coordinate planes: The projection of P(x,y,z) on the xy-, yz-, zx-planes are (x,y,0), (0,y,z), (x,0,z) respectively. These projections help visualize components of the position vector.

Worked numerical examples:

• P(2, -1, 3): position vector r = 2i - j + 3k; |r| = sqrt(2^2 + (-1)^2 + 3^2) = sqrt(14); unit vector u = (1/sqrt(14))(2i - j + 3k).
• P(1,2,3), Q(4,0,-1): PQ = (3, -2, -4); distance = sqrt(3^2 + (-2)^2 + (-4)^2) = sqrt(29); midpoint = (2.5, 1, 1).

Why this matters / uses: Position vectors provide a compact vector description of location, simplify geometry and physics calculations (distances, forces, motion), and are foundational in 3D modelling, robotics, navigation and engineering design.

Definition — Direction Ratios (DR)
If a straight line (or a vector) in 3‑D has components proportional to (a, b, c), then a, b, c are called its direction ratios. Any non‑zero scalar multiple ka, kb, kc (k ≠ 0) represents the same direction ratios.

Definition — Direction Cosines (DC)
If α, β, γ are the angles made by a line (or a vector) with the positive x, y and z axes respectively, the direction cosines are l = cosα, m = cosβ and n = cosγ. The triple (l, m, n) are the direction cosines and they satisfy l² + m² + n² = 1.

Relation between DR and DC
If (a, b, c) are direction ratios of a line, its direction cosines are l = a / √(a² + b² + c²), m = b / √(a² + b² + c²), n = c / √(a² + b² + c²). Conversely, direction ratios can be taken as (l, m, n) multiplied by any nonzero constant.

Lines in 3D using direction ratios/cosines
Line through (x1, y1, z1) with direction ratios (a, b, c): (x − x1)/a = (y − y1)/b = (z − z1)/c (symmetric form). Parametric form: x = x1 + at, y = y1 + bt, z = z1 + ct.

Angle between two lines
If two lines have direction cosines (l1, m1, n1) and (l2, m2, n2), the cosine of the angle θ between them is cos θ = l1l2 + m1m2 + n1n2. If given direction ratios (a1, b1, c1) and (a2, b2, c2), use cos θ = (a1a2 + b1b2 + c1c2) / (√(a1²+b1²+c1²) · √(a2²+b2²+c2²)).

Perpendicular and parallel conditions
Two lines are perpendicular iff a1a2 + b1b2 + c1c2 = 0 (or l1l2 + m1m2 + n1n2 = 0). They are parallel iff their direction ratios are proportional.

Notes
• Direction cosines are signed; knowing two determines the third up to sign: n = ±√(1 − l² − m²).
• For plane normals: coefficients (A, B, C) of plane Ax + By + Cz + D = 0 are direction ratios of the normal; direction cosines of normal are A/√(A²+B²+C²), etc.

Overview. Vector algebra in 3D uses vectors (directed line segments) in three-dimensional space to model positions, directions, magnitudes and relationships between geometric objects. It provides efficient methods to represent lines, planes, angles, distances, areas and volumes using algebraic operations: addition, scalar multiplication, dot product and cross product.

Vectors and position vectors. A vector is represented as v = (v_x, v_y, v_z). The position vector of a point P(x,y,z) from the origin O is r = xi + yj + zk. Operations (addition and scalar multiplication) are coordinate-wise.

Dot product (scalar product). For a = (a_x,a_y,a_z) and b = (b_x,b_y,b_z), a·b = a_x b_x + a_y b_y + a_z b_z = |a||b| cosθ. Use it to find the angle between vectors and projections.

Cross product (vector product). a × b = (a_y b_z − a_z b_y, a_z b_x − a_x b_z, a_x b_y − a_y b_x). The result is perpendicular to both a and b, with magnitude |a × b| = |a||b| sinθ. Its magnitude equals the area of the parallelogram formed by a and b.

Projections. The vector projection of a on b: proj_b(a) = ((a·b)/|b|^2) b. Scalar projection = (a·b)/|b|.

Scalar triple product and coplanarity. a·(b × c) gives the signed volume of the parallelepiped formed by a, b, c. If a·(b × c) = 0, the three vectors are coplanar.

Key geometric applications.

• Equation of a line (vector form): r = a + t b, where a is a point on the line and b is a direction vector.
• Equation of a plane (vector form): (r − r0)·n = 0 or r·n = d, where n is a normal vector to the plane and r0 is a known point.
• Angle between lines/planes: Use dot product of direction vectors / normals: cosθ = (u·v)/(|u||v|).
• Distance from point to line: Distance = |(r_p − a) × b| / |b|, where a is any point on the line and b is its direction vector.
• Distance from point to plane: Distance = |(r_p − r0)·n| / |n|.
• Area of triangle ABC: (1/2)|AB × AC|. Area of parallelogram = |AB × AC|.
• Volume of parallelepiped: |a·(b × c)|.

Solving common problems — quick steps.

• To get line through two points P and Q: direction = Q − P; use r = P + t(Q − P).
• To get plane through point P with two direction vectors u and v: normal n = u × v; plane: (r − P)·n = 0.
• Intersection of two planes: direction = n1 × n2; find a point on both planes (solve two linear equations) then r = p + t(n1 × n2).
• Shortest distance and perpendicular foot: compute projection using dot/cross formulas; for a point to plane, foot = r_p − ((r_p − r0)·n)/|n|^2 · n.

Useful vector identities (helpful for derivations): a × b = −(b × a); a·(b × c) is invariant under cyclic permutations; a × (b × c) = b(a·c) − c(a·b).

Tips for students. Always identify position vectors and direction/normal vectors. Use cross product to get normals or directions perpendicular to two given vectors. Use dot product to get angles and projections. Keep track of magnitudes for distances and areas.

Introduction: In three-dimensional (3D) geometry a straight line is the set of points that extends infinitely in two opposite directions and lies along a fixed direction. A line can be specified by a point it passes through and its direction. There are several equivalent algebraic forms used in Class 12: vector form, parametric form, symmetric (Cartesian) form, and two-plane (intersection) form.

1. Vector form: If a line passes through a fixed point A with position vector a and has direction vector b, any point P on the line has position vector r given by

r = a + λ b, λ ∈ ℝ.

This is compact and useful for vector operations (dot, cross, projection, distances).

2. Parametric form: Writing a = (x1,y1,z1) and b = (l,m,n), the vector equation gives parametric equations

x = x1 + λ l, y = y1 + λ m, z = z1 + λ n.

3. Symmetric (Cartesian) form: If l,m,n are all nonzero, eliminate λ to get

(x - x1)/l = (y - y1)/m = (z - z1)/n.

Special note: if a direction component is zero (say n = 0), then z = z1 and the symmetric form uses the remaining equalities: (x-x1)/l = (y-y1)/m, z = z1.

4. Two-point form: A line through two distinct points A(x1,y1,z1) and B(x2,y2,z2) has direction ratios (x2-x1, y2-y1, z2-z1). Parametric or symmetric forms follow by substituting these as l,m,n.

5. Line as intersection of two planes: The intersection of two non-parallel planes P1: u1 x + v1 y + w1 z + d1 = 0 and P2: u2 x + v2 y + w2 z + d2 = 0 is a straight line. Solve the two plane equations simultaneously to get parametric equations of that line (choose a free parameter or solve for two variables in terms of the third).

Key geometric relations:

• Direction ratios (DRs) of a line: l : m : n. Direction cosines α, β, γ satisfy α² + β² + γ² = 1 and are proportional to l,m,n.
• Angle θ between two lines with direction vectors b1 = (l1,m1,n1) and b2 = (l2,m2,n2): cos θ = (l1 l2 + m1 m2 + n1 n2) / (|b1| |b2|).
• Parallelism: two lines are parallel if their direction vectors are scalar multiples. Perpendicularity: dot product = 0.

Distances (useful in many problems):

• Distance from a point P(x0,y0,z0) to line through A(x1,y1,z1) with direction vector b: distance = |(AP × b)| / |b| where AP = (x0-x1,y0-y1,z0-z1).
• Shortest distance between two skew lines r = a1 + λ b1 and r = a2 + μ b2: distance = |(b1 × b2) · (a2 - a1)| / |b1 × b2|.

Worked idea: To write the equation of the line through points A(1,2,0) and B(4,0,3), take direction vector b = B - A = (3,-2,3). Vector form: r = (1,2,0) + λ(3,-2,3). Parametric: x = 1 + 3λ, y = 2 - 2λ, z = 0 + 3λ. Symmetric: (x-1)/3 = (y-2)/-2 = z/3.

Why this matters / intuitive uses: Representing a straight element in space (beam, sight-line, flight path) in analytic form lets you compute intersections, shortest distances, angles, and projections algebraically.

Tips for solving Class 12 problems:

• Choose a convenient parameter (λ or t). For intersection with a plane substitute parametrics into plane equation to find parameter value(s).
• When two components of the direction vector are zero treat the remaining coordinate as constant (line parallel to an axis).
• Use vector cross product formulas for distances and shortest segments between skew lines.

Definition. The distance from a point P to a line l in space is the length of the perpendicular segment from P to l — i.e. the shortest distance between the point and any point on the line.

Set-up (vector form). Let the line l be given in vector form r = a + t b, where a is a position vector of a point A on the line and b is a direction vector of the line. Let P be the given point with position vector r_0. Put AP = r_0 - a.

Derivation. Consider the parallelogram made by vectors b and AP. Its area equals |b x AP|. But the same area equals |b| |AP| sinθ, where θ is the angle between b and AP. The perpendicular distance d from P to the line is |AP| sinθ. Thus

d = |b x (r_0 - a)| / |b|.

This is the standard 3D formula. Geometrically, b x (r_0 - a) is a vector whose magnitude equals the area of the parallelogram; dividing by |b| leaves the height (perpendicular distance).

Foot of perpendicular. The foot H of the perpendicular from P to the line has parameter t_0 = (b · (r_0 - a)) / |b|^2. The coordinates (position vector) of H are a + t_0 b. If you need the distance to a line segment AB rather than the infinite line, first compute t_0 relative to A and B: if t_0 lies between 0 and 1 the perpendicular hits the segment; otherwise the shortest distance is to the nearer endpoint.

Component / coordinate forms.

• If the line is given in symmetric form (x - x1)/l = (y - y1)/m = (z - z1)/n with direction ratios l,m,n, and P has coordinates (x0,y0,z0), the distance can be written using components as

d = sqrt( ((y1 - y0) n - (z1 - z0) m)^2 + ((z1 - z0) l - (x1 - x0) n)^2 + ((x1 - x0) m - (y1 - y0) l)^2 ) / sqrt(l^2 + m^2 + n^2).

2D special case. For a line ax + by + c = 0 in the plane and point (x0,y0) the well-known formula is

d = |a x0 + b y0 + c| / sqrt(a^2 + b^2).

Remarks. The vector cross-product formula is compact and practical in 3D. Always check whether you want distance to an infinite line or to a finite segment — for segments the perpendicular foot may fall outside the segment.

Definition. Two lines in space are called skew if they are neither parallel nor intersecting. Skew lines lie in different planes.

Vector/Parametric form of a line. A line L through point a (position vector a) with direction vector b is written as r = a + λ b, λ ∈ ℝ.

When are two lines parallel, intersecting or skew?

• Given L1: r = a1 + λ b1 and L2: r = a2 + μ b2,
• Parallel ⇔ b1 × b2 = 0 (direction vectors are collinear).
• Intersect ⇔ there exist λ, μ such that a1 + λ b1 = a2 + μ b2. Equivalently, if b1 × b2 ≠ 0 and (a2 − a1) · (b1 × b2) = 0 then the lines meet (they lie in a common plane).
• Skew ⇔ b1 × b2 ≠ 0 and (a2 − a1) · (b1 × b2) ≠ 0.

Shortest distance between two skew lines (geometric idea). The shortest segment joining two skew lines is the line segment that is perpendicular to both direction vectors b1 and b2. Its direction is parallel to n = b1 × b2 (the normal to both directions).

Formula (scalar triple product). If L1: r = a1 + λ b1 and L2: r = a2 + μ b2 and b1 × b2 ≠ 0, then the shortest distance d between the two lines is

d = |(a2 − a1) · (b1 × b2)| / |b1 × b2|.

How to find the actual closest points (common perpendicular). Solve for λ and μ from the two linear equations ensuring the connector vector is orthogonal to both directions:

(a1 − a2 + λ b1 − μ b2) · b1 = 0
(a1 − a2 + λ b1 − μ b2) · b2 = 0.

Tip. If the numerator in the distance formula is zero and b1 × b2 ≠ 0, the lines intersect (distance zero).

Definition: A plane in 3-dimensional space is a flat two-dimensional surface that extends infinitely. A plane can be uniquely determined by (i) a point and a normal vector, (ii) three non-collinear points, or (iii) a point and two independent direction vectors lying in the plane.

Vector (normal) / Cartesian form: Let n = (a, b, c) be a normal vector to the plane and let P0(x0, y0, z0) be a point on it. For any point P(x, y, z) in the plane, the vector P0P is perpendicular to n, so n . P0P = 0. This gives the point–normal form:

a(x - x0) + b(y - y0) + c(z - z0) = 0.

Expanding and writing d = -(ax0 + by0 + cz0) yields the general (Cartesian) equation of a plane:

ax + by + cz + d = 0, where (a, b, c) is the normal vector.

Parametric / Vector form: If r0 is the position vector of a point on the plane and u, v are two non-parallel direction vectors lying in the plane, then every point r on the plane can be written as

r = r0 + s u + t v, for real parameters s, t.

The normal vector can be obtained as n = u x v (cross product).

Intercept form (when plane meets axes at a, b, c nonzero):

x/a + y/b + z/c = 1.

Other important facts:

• Angle between two planes with normals n1 and n2: cos theta = |n1 . n2| / (|n1| |n2|). (Use absolute if only the acute angle is required.)
• Distance from point Q(x1, y1, z1) to plane ax + by + cz + d = 0: distance = |ax1 + by1 + cz1 + d| / sqrt(a^2 + b^2 + c^2).
• Plane through intersection of two planes P1: a1x + b1y + c1z + d1 = 0 and P2: a2x + b2y + c2z + d2 = 0 is: P1 + λ P2 = 0 (family of planes), parameter λ gives different planes passing through the line of intersection.
• Special planes: x = const (plane parallel to yz-plane), y = const, z = const.

How to get a plane from three points: Given A, B, C (non-collinear), two direction vectors are AB and AC. Compute n = AB x AC, then use point–normal form with any of the three points.

Overview. In 3‑D geometry, angles are measured between directions (vectors). The angle between two lines is the angle between their direction vectors. The angle between a line and a plane is the complement of the angle between the line's direction vector and the plane's normal. The angle between two planes is the angle between their normals.

Angle between two lines. If two lines have direction vectors b and d, the angle θ between the lines is given by

θ = arccos[(b · d) / (|b| |d|)].

If the acute angle is required, use θ = arccos(|b · d| / (|b| |d|)). This formula applies whether the lines intersect or are skew: it uses their direction vectors.

Angle between a line and a plane. Let a line have direction vector b and let the plane have unit normal n (or normal vector n). Let α be the angle between the line and the plane. If β is the angle between the line and the normal, then α + β = 90°.

β = arccos[(|b · n|) / (|b| |n|)], so
α = 90° − β = arcsin[(|b · n|) / (|b| |n|)].

Equivalently, sin(α) = |b · n| / (|b||n|). The absolute value ensures we take the acute angle between the line and the plane.

Angle between two planes. If planes have normals n1 and n2, the angle φ between the planes is the angle between n1 and n2:

φ = arccos[(n1 · n2) / (|n1| |n2|)].

Again, to get the acute angle you may take the absolute value of the dot product.

Notes on signs and intersections. When two planes intersect, their line of intersection lies at the intersection of the planes; the angle between planes is measured by the angle between their normals (or its supplement). For skew lines (non-parallel, non-intersecting), use their direction vectors as usual.

Concept

Given a plane with equation ax + by + cz + d = 0 and a point P(x1, y1, z1), the shortest (perpendicular) distance from P to the plane is the length of the projection of the vector from any point on the plane to P along the plane's normal vector n = (a, b, c). The distance is always nonnegative and is obtained along the direction of the normal.

Formula (derived by projection)

Distance = |a x1 + b y1 + c z1 + d| / sqrt(a^2 + b^2 + c^2).

Foot of the perpendicular

If H(x0, y0, z0) is the foot of perpendicular from P(x1, y1, z1) to the plane ax + by + cz + d = 0, then

x0 = x1 - a * (a x1 + b y1 + c z1 + d) / (a^2 + b^2 + c^2)

y0 = y1 - b * (a x1 + b y1 + c z1 + d) / (a^2 + b^2 + c^2)

z0 = z1 - c * (a x1 + b y1 + c z1 + d) / (a^2 + b^2 + c^2)

Distance between two parallel planes

Let two parallel planes be written as ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 (same normal vector n = (a,b,c)). The shortest distance between them is

Distance = |d1 - d2| / sqrt(a^2 + b^2 + c^2).

Remarks

• The numerator for distance from a point to a plane gives a signed value; taking absolute value gives the actual distance. The sign indicates which side of the plane the point lies on relative to the direction of n.
• Make sure the planes are expressed with the same normal coefficients (a, b, c) before using the formula for distance between planes.

Overview. In three-dimensional geometry, a line can intersect a plane at a point, be parallel to it (no intersection), or lie entirely in the plane (infinitely many intersections). Understanding intersections is essential for finding shortest distances, foots of perpendiculars, and for applications in modeling, engineering and graphics.

Representations.

• Line (vector/parametric): r = r0 + t v, where r0 = (x0,y0,z0) is a point on the line, v = (l,m,n) is the direction vector, t ∈ R.
• Plane (Cartesian): ax + by + cz + d = 0, with normal vector n = (a,b,c).

How to find intersection (method).

• Substitute the parametric coordinates of the line (x,y,z) = (x0 + lt, y0 + mt, z0 + nt) into the plane equation ax+by+cz+d=0.
• This yields a linear equation in t: a(x0+lt)+b(y0+mt)+c(z0+nt)+d = 0. Solve for t.
• If there is a unique t, the line meets the plane at the point r0 + t v.
• If the equation is inconsistent (no t), the line is parallel to the plane and not in it.
• If the equation holds for all t, every point of the line satisfies the plane equation, so the line lies in the plane.

Special conditions.

• Line parallel to plane: v · n = al + bm + cn = 0. If additionally a x0 + b y0 + c z0 + d = 0, the line lies in the plane; otherwise it is parallel and distinct.
• Angle between line and plane: If θ is the angle between the line (direction v) and the plane, and φ is the angle between v and the plane's normal n, then θ = 90° − φ. Useful relation: sin(θ) = |v · n|/(|v||n|).

Related tasks (applications).

• Foot of perpendicular from a point to a plane (closest point on plane).
• Shortest distance from a point to a plane (used in collision detection, error minimization).
• Intersection line of two planes (used in structural lines, edges of polyhedra).
• Finding where a ray hits a surface (computer graphics, ray-tracing).

Summary procedure (line vs plane). Given line r = r0 + t v and plane ax + by + cz + d = 0, substitute to get a linear equation in t. Solve it. Use v·n to detect parallelism or containment.