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Vector Algebra Formulas For JEE 2027
Vector Algebra in JEE Mathematics deals with quantities that have both magnitude and direction. The chapter covers vector addition, scalar multiplication, components, unit vectors, and the dot and cross products. These concepts help students calculate angles between vectors, find projections, test perpendicularity or parallelism, and determine areas of triangles and parallelograms.
Effective preparation starts with understanding each operation and recognising when to use it. Practise problems in both component and geometric forms, paying attention to signs, vector directions, and the order of cross products. Use a JEE Mains formula PDF to revise key results, then apply them through chapter-wise questions. Taking a JEE Mains mock test can help you assess your recall, identify mistakes, and practise solving Vector Algebra questions under time limits.
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 2027: Conclusion
Preparing well in vector algebra means understanding what each operation represents and when to apply it. Revise vector properties, projections, dot products, and cross products, then solve questions involving angles, areas, and geometric relationships. Use a JEE Mains Maths Formulas sheet to review key results, paying particular attention to signs, directions, and the order of cross products.
As you plan your revision around JEE Mains Maths Important Chapters, include regular Vector Algebra practice sessions. Solve previous-year questions, review incorrect attempts, and practise under time limits to identify where you lose time or make calculation errors. Consistent revision and careful application of concepts can help you approach this chapter with greater accuracy and confidence.
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