Vector Algebra Formulas For JEE 2026
Vector Algebra is an important chapter in JEE Mains Mathematics and usually carries 1–2 questions in the exam. In this topic, students learn the basics of vectors, including operations like addition and scalar multiplication. It also covers key concepts such as the dot product and cross product, which are useful for solving both algebraic and geometric problems.
The chapter further includes topics like the scalar triple product and vector triple product, along with their applications in 3D geometry. Since many questions are based on formulas, having a clear understanding and practicing regularly is very important. For quick revision before exams, students can also use a well-organized JEE Mains Maths Formula PDF to revise important formulas and concepts easily.
Vectors: Basics and Types
Position Vector
The vector from the origin $$O$$ to a point $$P$$ is called the position vector of $$P$$. If $$P = (x, y, z)$$, then $$\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$$.
Magnitude of a Vector
If $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$, then:
$$|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$$Special Types of Vectors
- Zero vector ($$\vec{0}$$): Magnitude 0, no definite direction.
- Unit vector ($$\hat{a}$$): Magnitude 1. $$\hat{a} = \dfrac{\vec{a}}{|\vec{a}|}$$
- Equal vectors: Same magnitude and direction.
- Negative vector: $$-\vec{a}$$ has same magnitude, opposite direction.
- Collinear vectors: Parallel vectors. $$\vec{a} \parallel \vec{b}$$ means $$\vec{a} = \lambda \vec{b}$$.
- Coplanar vectors: Vectors that lie in the same plane.
Worked Example
Find the unit vector in the direction of $$\vec{a} = 2\hat{i} + 3\hat{j} - 6\hat{k}$$.
$$|\vec{a}| = \sqrt{4 + 9 + 36} = 7$$
$$\hat{a} = \dfrac{2}{7}\hat{i} + \dfrac{3}{7}\hat{j} - \dfrac{6}{7}\hat{k}$$
Addition of Vectors
Component-wise Addition
If $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$ and $$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$$:
$$\vec{a} + \vec{b} = (a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j} + (a_3 + b_3)\hat{k}$$ $$\vec{a} - \vec{b} = (a_1 - b_1)\hat{i} + (a_2 - b_2)\hat{j} + (a_3 - b_3)\hat{k}$$Scalar Multiplication
If $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$ and $$\lambda$$ is a scalar:
$$\lambda\vec{a} = \lambda a_1 \hat{i} + \lambda a_2 \hat{j} + \lambda a_3 \hat{k}, \quad |\lambda\vec{a}| = |\lambda| \cdot |\vec{a}|$$Section Formula
If $$A$$ and $$B$$ have position vectors $$\vec{a}$$ and $$\vec{b}$$, and point $$P$$ divides $$AB$$:
Internally in ratio $$m : n$$: $$\vec{p} = \dfrac{m\vec{b} + n\vec{a}}{m + n}$$
Externally in ratio $$m : n$$: $$\vec{p} = \dfrac{m\vec{b} - n\vec{a}}{m - n}$$
Midpoint: $$\vec{p} = \dfrac{\vec{a} + \vec{b}}{2}$$
Dot Product (Scalar Product)
For two vectors $$\vec{a}$$ and $$\vec{b}$$ with angle $$\theta$$ between them: $$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$$
Dot Product in Component Form
$$\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$Properties of Dot Product
- Commutative: $$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$$
- Distributive: $$\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$$
- $$\vec{a} \cdot \vec{a} = |\vec{a}|^2$$
- If $$\vec{a} \perp \vec{b}$$, then $$\vec{a} \cdot \vec{b} = 0$$
- If $$\vec{a} \parallel \vec{b}$$, then $$\vec{a} \cdot \vec{b} = \pm|\vec{a}||\vec{b}|$$
- $$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$$; $$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$$
Angle Between Two Vectors
$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{a_1 b_1 + a_2 b_2 + a_3 b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\;\sqrt{b_1^2+b_2^2+b_3^2}}$$Worked Example
Find the angle between $$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$$ and $$\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$$.
$$\vec{a} \cdot \vec{b} = 2 - 6 + 2 = -2$$. $$|\vec{a}| = \sqrt{14}$$, $$|\vec{b}| = 3$$.
$$\cos\theta = \dfrac{-2}{3\sqrt{14}}$$, $$\theta = \cos^{-1}\!\left(\dfrac{-2}{3\sqrt{14}}\right)$$
Projection of a Vector
Projection Formulas
Scalar projection of $$\vec{a}$$ on $$\vec{b}$$:
$$\text{proj} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$Vector projection of $$\vec{a}$$ on $$\vec{b}$$:
$$\text{proj}_{\vec{b}}\,\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\;\vec{b}$$Worked Example
Find the projection of $$\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$$ on $$\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$$.
$$\vec{a} \cdot \vec{b} = 6 - 2 - 1 = 3$$. $$|\vec{b}| = 3$$.
Scalar projection $$= \dfrac{3}{3} = \textbf{1}$$
Tip: If the projection is positive, $$\vec{a}$$ has a component in the direction of $$\vec{b}$$. If negative, opposite to $$\vec{b}$$. If zero, the vectors are perpendicular.
Cross Product (Vector Product)
$$\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\;\hat{n}$$, where $$\hat{n}$$ is the unit vector perpendicular to both $$\vec{a}$$ and $$\vec{b}$$ (right-hand rule).
Cross Product in Component Form
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\hat{i} - (a_1 b_3 - a_3 b_1)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k}$$Properties of Cross Product
- Anti-commutative: $$\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$$
- Distributive: $$\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$$
- NOT associative in general
- $$\vec{a} \times \vec{a} = \vec{0}$$
- If $$\vec{a} \parallel \vec{b}$$, then $$\vec{a} \times \vec{b} = \vec{0}$$
- $$\hat{i} \times \hat{j} = \hat{k}$$, $$\hat{j} \times \hat{k} = \hat{i}$$, $$\hat{k} \times \hat{i} = \hat{j}$$
Worked Example
Find $$\vec{a} \times \vec{b}$$ if $$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$$ and $$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$$.
$$\vec{a} \times \vec{b} = \hat{i}(6 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 - 3) = \boldsymbol{7\hat{i} - 3\hat{j} - 5\hat{k}}$$
Area Using Cross Product
Area Formulas
- Area of parallelogram with adjacent sides $$\vec{a}$$ and $$\vec{b}$$: $$\text{Area} = |\vec{a} \times \vec{b}|$$
- Area of triangle with sides $$\vec{a}$$ and $$\vec{b}$$: $$\text{Area} = \dfrac{1}{2}|\vec{a} \times \vec{b}|$$
- Area of parallelogram with diagonals $$\vec{d_1}$$ and $$\vec{d_2}$$: $$\text{Area} = \dfrac{1}{2}|\vec{d_1} \times \vec{d_2}|$$
Worked Example
Find the area of the triangle with vertices $$A(1,1,1)$$, $$B(2,3,1)$$, $$C(1,2,3)$$.
$$\vec{AB} = \hat{i} + 2\hat{j}$$, $$\vec{AC} = \hat{j} + 2\hat{k}$$
$$\vec{AB} \times \vec{AC} = 4\hat{i} - 2\hat{j} + \hat{k}$$, $$|\vec{AB} \times \vec{AC}| = \sqrt{21}$$
Area $$= \dfrac{1}{2}\sqrt{21}$$ square units
Scalar Triple Product
Scalar Triple Product Determinant Form
$$[\vec{a}\;\vec{b}\;\vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$Properties of Scalar Triple Product
- Cyclic permutation: $$[\vec{a}\;\vec{b}\;\vec{c}] = [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{c}\;\vec{a}\;\vec{b}]$$
- Sign change on swap: $$[\vec{a}\;\vec{b}\;\vec{c}] = -[\vec{b}\;\vec{a}\;\vec{c}]$$
- Volume of parallelepiped $$= |[\vec{a}\;\vec{b}\;\vec{c}]|$$
- Volume of tetrahedron $$= \dfrac{1}{6}|[\vec{a}\;\vec{b}\;\vec{c}]|$$
- Coplanarity condition: $$[\vec{a}\;\vec{b}\;\vec{c}] = 0$$
Worked Example
Find the volume of the parallelepiped formed by $$\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$$, $$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$$, $$\vec{c} = 3\hat{i} - \hat{j} + \hat{k}$$.
$$[\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 2 & -3 & 1 \\ 1 & 1 & -2 \\ 3 & -1 & 1 \end{vmatrix} = 2(-1) + 3(7) + (-4) = 15$$
Volume $$= |15| = \textbf{15}$$ cubic units
Tip: Cyclic order preserves the sign, swapping any two vectors changes the sign. If you get 0, the vectors are coplanar.
Vector Triple Product
Vector Triple Product (BAC-CAB Rule)
$$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$ $$(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$$Key point: $$\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}$$ in general. The result always lies in the plane of the two vectors inside the brackets.
Worked Example
If $$\vec{a} = \hat{i} + \hat{j}$$, $$\vec{b} = \hat{j} + \hat{k}$$, $$\vec{c} = \hat{k} + \hat{i}$$, find $$\vec{a} \times (\vec{b} \times \vec{c})$$.
$$\vec{a} \cdot \vec{c} = 1$$, $$\vec{a} \cdot \vec{b} = 1$$
$$\vec{a} \times (\vec{b} \times \vec{c}) = 1 \cdot (\hat{j} + \hat{k}) - 1 \cdot (\hat{k} + \hat{i}) = \boldsymbol{-\hat{i} + \hat{j}}$$
Tip: Remember BAC-CAB: $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$$. The vector outside the bracket dots with the vectors inside.
Quick Reference: Key Formulas
Summary Table
| Operation | Formula | Result Type |
|---|---|---|
| Magnitude | $$|\vec{a}| = \sqrt{a_1^2+a_2^2+a_3^2}$$ | Scalar |
| Dot product | $$\vec{a}\cdot\vec{b} = a_1b_1+a_2b_2+a_3b_3$$ | Scalar |
| Cross product | Determinant form | Vector |
| Triple scalar | $$[\vec{a}\;\vec{b}\;\vec{c}]$$ = determinant | Scalar |
| Angle | $$\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$$ | Scalar |
| Area (triangle) | $$\dfrac{1}{2}|\vec{a}\times\vec{b}|$$ | Scalar |
| Volume (parallelepiped) | $$|[\vec{a}\;\vec{b}\;\vec{c}]|$$ | Scalar |
Vector Algebra Formulas For JEE 2026: Conclusion
Vector algebra is one of the most important and scoring topics in mathematics for competitive exams. Concepts like dot product, cross product, and vector operations help in solving both algebraic and geometric problems efficiently. With a clear understanding of formulas and regular practice, students can easily improve their accuracy and speed in this chapter.
To perform well, it is essential to revise key concepts like projections, triple products, and vector properties consistently. Practicing different types of problems and focusing on applications will strengthen problem-solving skills. With the right preparation strategy, this chapter can become a strong area for scoring marks in the exam.
