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Vectors & 3D Geometry Formulas

Complete chapter-wise formula sheet for Vectors & 3D Geometry (Mathematics) covering key equations, parameters, traps, and concept diagrams for JEE Main & Advanced.

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Chapter Directory (26 Chapters)

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Vectors & 3D Geometry1,174 Qs

Three Dimensional Geometry (3D)

Direction cosines & ratios, equations of lines in 3D, shortest distance between skew and parallel lines.

9 Formulas Diagram
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Vectors & 3D Geometry1,073 Qs

Vector Algebra

Dot product, cross product, scalar triple product, vector triple product, and Lagrange's identity.

7 Formulas Diagram
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Formulas Reference (Vectors & 3D Geometry)

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Vectors & 3D GeometryTop Priority (3-4 Qs in JEE)1,174 PYQs

Three Dimensional Geometry (3D)

Direction cosines & ratios, equations of lines in 3D, shortest distance between skew and parallel lines.

Chapter

Direction Cosines & Direction Ratios

l=cos⁡α,  m=cos⁡β,  n=cos⁡γ  ⟹  l2+m2+n2=1,l=aa2+b2+c2l = \cos\alpha, \; m = \cos\beta, \; n = \cos\gamma \implies l^2 + m^2 + n^2 = 1, \quad l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}

Direction cosines represent components of the unit direction vector in 3D.

cos²α + cos²β + cos²γ = 1 implies sin²α + sin²β + sin²γ = 2.

Equation of a Line in 3D

x−x1a=y−y1b=z−z1c=λ  ⟺  r⃗=a⃗+λb⃗\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda \iff \vec{r} = \vec{a} + \lambda \vec{b}

Symmetric and vector forms of line passing through (x₁, y₁, z₁) with direction ratios <a, b, c>.

Any general point on the line is (x₁ + aλ, y₁ + bλ, z₁ + cλ).

Shortest Distance Between Skew Lines

d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|

Minimum distance between non-parallel, non-intersecting skew lines in 3D.

Lines intersect if and only if (a₂ - a₁) · (b₁ × b₂) = 0.

Shortest Distance Between Parallel Lines

d=∣(a⃗2−a⃗1)×b⃗∣∣b⃗∣d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}

Distance between parallel lines r = a₁ + λb and r = a₂ + μb.

Uses cross product magnitude divided by |b|.

Angle Between Two Lines in 3D

cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12a22+b22+c22=∣b⃗1⋅b⃗2∣∣b⃗1∣∣b⃗2∣\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}} = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|}

Acute angle between lines with direction ratios.

Lines are perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0.

Condition for Coplanarity of Two Lines

∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣=0  ⟺  (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=0\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \iff (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0

Criterion for two straight lines in 3D space to lie in the same unique plane.

If coplanar, either the lines intersect at a point or they are parallel.

Distance of a Point from a Line in 3D

d=∣(p⃗−a⃗)×b⃗∣∣b⃗∣=∣p⃗−a⃗∣2−((p⃗−a⃗)⋅b⃗∣b⃗∣)2d = \frac{|(\vec{p} - \vec{a}) \times \vec{b}|}{|\vec{b}|} = \sqrt{|\vec{p} - \vec{a}|^2 - \left(\frac{(\vec{p} - \vec{a}) \cdot \vec{b}}{|\vec{b}|}\right)^2}

Perpendicular distance from point P(p) to line r = a + λb.

Foot of perpendicular F: let F = a + λb, then (p - F) · b = 0 determines λ directly.

Equation of Sphere & Its Properties

x2+y2+z2+2ux+2vy+2wz+d=0  ⟹  Centre=(−u,−v,−w),  r=u2+v2+w2−dx^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 \implies \text{Centre} = (-u,-v,-w), \; r = \sqrt{u^2 + v^2 + w^2 - d}

General second-degree equation of a sphere. Standard form with centre (a,b,c): (x-a)² + (y-b)² + (z-c)² = r².

Sphere through origin: d = 0. Diameter-form sphere with endpoints (x₁,y₁,z₁) and (x₂,y₂,z₂): (x-x₁)(x-x₂) + (y-y₁)(y-y₂) + (z-z₁)(z-z₂) = 0. For a point to lie on sphere: substitute and check if the equation is satisfied.

Image of a Point in a Plane & Foot of Perpendicular

x′−x1a=y′−y1b=z′−z1c=−2⋅ax1+by1+cz1+da2+b2+c2\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = \frac{z' - z_1}{c} = -2\cdot\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}

Coordinates of image (x', y', z') of point (x₁, y₁, z₁) in plane ax + by + cz + d = 0. Foot of perpendicular F has parameter k = −(ax₁+by₁+cz₁+d)/(a²+b²+c²).

F (foot of ⊥) = (x₁ + ak, y₁ + bk, z₁ + ck). Image point = (x₁ + 2ak, y₁ + 2bk, z₁ + 2ck). Distance from point to plane: |ax₁+by₁+cz₁+d| / √(a²+b²+c²).
Vectors & 3D GeometryTop Priority (3-4 Qs in JEE)1,073 PYQs

Vector Algebra

Dot product, cross product, scalar triple product, vector triple product, and Lagrange's identity.

Chapter

Scalar (Dot) Product & Projection

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ,Projection of a⃗ on b⃗=a⃗⋅b⃗∣b⃗∣\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta, \quad \text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Dot product yields scalar; zero if vectors are orthogonal.

Projection vector is ((a · b) / |b|²) b.

Vector (Cross) Product

a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ n^,Area of Parallelogram=∣a⃗×b⃗∣\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin\theta \, \hat{n}, \quad \text{Area of Parallelogram} = |\vec{a} \times \vec{b}|

Perpendicular vector directed by right-hand thumb rule.

Area of triangle with adjacent vectors a and b is (1/2)|a × b|.

Scalar Triple Product (Box Product)

[a⃗  b⃗  c⃗]=a⃗⋅(b⃗×c⃗)=∣a1a2a3b1b2b3c1c2c3∣[\vec{a} \; \vec{b} \; \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

Represents the volume of parallelepiped formed by coterminous vectors a, b, c.

Coplanarity condition: [a b c] = 0.

Vector Triple Product (BAC-CAB Rule)

a⃗×(b⃗×c⃗)=(a⃗⋅c⃗)b⃗−(a⃗⋅b⃗)c⃗\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}

Expands double cross product into a linear combination of vectors b and c.

Memory trick: BAC minus CAB.

Lagrange's Vector Identity

∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2

Connects cross product and dot product magnitudes directly.

Derived from sin²θ + cos²θ = 1.

Scalar & Vector Product of Four Vectors

(a⃗×b⃗)⋅(c⃗×d⃗)=∣a⃗⋅c⃗a⃗⋅d⃗b⃗⋅c⃗b⃗⋅d⃗∣,(a⃗×b⃗)×(c⃗×d⃗)=[a⃗  b⃗  d⃗]c⃗−[a⃗  b⃗  c⃗]d⃗(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \begin{vmatrix} \vec{a}\cdot\vec{c} & \vec{a}\cdot\vec{d} \\ \vec{b}\cdot\vec{c} & \vec{b}\cdot\vec{d} \end{vmatrix}, \quad (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = [\vec{a}\;\vec{b}\;\vec{d}]\vec{c} - [\vec{a}\;\vec{b}\;\vec{c}]\vec{d}

Determinant expansion for dot product of two cross products, and linear resolution of cross product of two cross products.

Crucial identity for finding line of intersection of two planes defined by normal vectors.

Reciprocal System of Vectors & Tetrahedron Volume

a⃗′=b⃗×c⃗[a⃗ b⃗ c⃗],a⃗⋅a⃗′=1,  a⃗⋅b⃗′=0,Vtetrahedron=16∣[a⃗ b⃗ c⃗]∣\vec{a}' = \frac{\vec{b} \times \vec{c}}{[\vec{a}\,\vec{b}\,\vec{c}]}, \quad \vec{a} \cdot \vec{a}' = 1, \; \vec{a} \cdot \vec{b}' = 0, \quad V_{\text{tetrahedron}} = \frac{1}{6}|[\vec{a}\,\vec{b}\,\vec{c}]|

Reciprocal triad of vectors satisfying orthogonal dot relations, and volume of tetrahedron with coterminous edges a, b, c.

The scalar triple product of reciprocal system is [a' b' c'] = 1 / [a b c].