Component Form Operations

If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then:

Addition: $\vec{a} + \vec{b} = (a_1+b_1)\hat{i} + (a_2+b_2)\hat{j} + (a_3+b_3)\hat{k}$

Subtraction: $\vec{a} - \vec{b} = (a_1-b_1)\hat{i} + (a_2-b_2)\hat{j} + (a_3-b_3)\hat{k}$

Equal Vectors: $\vec{a} = \vec{b}$ if and only if $a_1 = b_1, a_2 = b_2, a_3 = b_3$

Vector Joining Two Points

Vector from $P_1(x_1, y_1, z_1)$ to $P_2(x_2, y_2, z_2)$: $\vec{P_1P_2} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$ $|\vec{P_1P_2}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Section Formula ★★★★★

Internal Division: If R divides PQ internally in ratio m:n, then: $\vec{OR} = \frac{m\vec{OQ} + n\vec{OP}}{m+n}$
External Division: If R divides PQ externally in ratio m:n, then: $\vec{OR} = \frac{m\vec{OQ} - n\vec{OP}}{m-n}$
Midpoint Formula: (when m = n = 1) $\vec{OR} = \frac{\vec{OP} + \vec{OQ}}{2}$
Memory Aid:
• Internal: "+" in numerator, "+" in denominator
• External: "−" in numerator, "−" in denominator
• For midpoint: Just average the position vectors!
Common Board Question Pattern:
Given three points A, B, C. Find position vector of point dividing AB in ratio 2:1.
Solution Method:
1. Write position vectors of A and B
2. Apply section formula: $\vec{r} = \frac{2\vec{b} + 1\vec{a}}{2+1}$
3. Simplify to get answer

Dot Product (Scalar Product) 🔥🔥🔥

Definition: The dot product of two vectors $\vec{a}$ and $\vec{b}$ is: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, $0 \leq \theta \leq \pi$
Component Form: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$

Properties of Dot Product

Property Formula
Commutative $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$
Distributive $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$
Scalar Multiple $(k\vec{a}) \cdot \vec{b} = k(\vec{a} \cdot \vec{b})$
Self Dot Product $\vec{a} \cdot \vec{a} = |\vec{a}|^2$
Unit Vectors $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$
Perpendicular $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

Important Results

1. Angle Between Vectors: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

2. Perpendicular Vectors: $\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$

3. Parallel Vectors: $\vec{a} \parallel \vec{b} \iff \vec{a} = k\vec{b} \text{ or } \theta = 0 \text{ or } \pi$
Quick Checks:
• If $\vec{a} \cdot \vec{b} = 0$ → vectors are perpendicular
• If $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|$ → vectors are parallel (same direction)
• If $\vec{a} \cdot \vec{b} = -|\vec{a}||\vec{b}|$ → vectors are parallel (opposite direction)

Projection of Vector

Projection of $\vec{a}$ on $\vec{b}$: $\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = |\vec{a}|\cos\theta$

Cross Product (Vector Product) ★★★★★

Definition: The cross product of two vectors $\vec{a}$ and $\vec{b}$ is: $\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\, \hat{n}$ where $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ (right-hand rule)
Component Form (Determinant Method): $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$ $= (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$

Properties of Cross Product

Property Formula
Not Commutative $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$
Distributive $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$
Scalar Multiple $k(\vec{a} \times \vec{b}) = (k\vec{a}) \times \vec{b} = \vec{a} \times (k\vec{b})$
Self Cross $\vec{a} \times \vec{a} = \vec{0}$
Unit Vectors $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$
Reverse Order $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{j} = -\hat{i}$, $\hat{i} \times \hat{k} = -\hat{j}$
Memory Trick for $\hat{i}, \hat{j}, \hat{k}$ Cross Products:
Cyclic order (clockwise): $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$
Anti-cyclic (reverse): Just add negative sign!

Geometric Applications

1. Area of Triangle with vertices A, B, C: $\text{Area} = \frac{1}{2}|\vec{AB} \times \vec{AC}|$

2. Area of Parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$: $\text{Area} = |\vec{a} \times \vec{b}|$

3. Unit Vector Perpendicular to both $\vec{a}$ and $\vec{b}$: $\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
Parallel Vectors: $\vec{a} \parallel \vec{b}$ if and only if $\vec{a} \times \vec{b} = \vec{0}$

Solved Examples (Board Marking Scheme)

Q1. Find the unit vector in the direction of $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ (2 Marks)

Step 1: Find Magnitude 1 Mark

$|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$

Step 2: Calculate Unit Vector 1 Mark

$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}}$

$= \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k}$

Q2. Find the angle between vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ (3 Marks)

Step 1: Calculate Dot Product 1 Mark

$\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1$

Step 2: Find Magnitudes 0.5 Mark

$|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

$|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$

Step 3: Apply Formula 1 Mark

$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{1}{\sqrt{3} \times \sqrt{3}} = \frac{1}{3}$

Step 4: Find Angle 0.5 Mark

$\theta = \cos^{-1}\left(\frac{1}{3}\right)$

Q3. Find $\vec{a} \times \vec{b}$ if $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}$ (3 Marks)

Solution: 3 Marks

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix}$

$= \hat{i}(1 \times (-2) - 3 \times 5) - \hat{j}(2 \times (-2) - 3 \times 3) + \hat{k}(2 \times 5 - 1 \times 3)$

$= \hat{i}(-2 - 15) - \hat{j}(-4 - 9) + \hat{k}(10 - 3)$

$= -17\hat{i} + 13\hat{j} + 7\hat{k}$

Q4. Find the area of triangle with vertices A(1, 1, 2), B(2, 3, 5), C(1, 5, 5) (5 Marks)

Step 1: Find Vectors 1 Mark

$\vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$

$\vec{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 4\hat{j} + 3\hat{k}$

Step 2: Calculate Cross Product 2 Marks

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix}$

$= \hat{i}(6-12) - \hat{j}(3-0) + \hat{k}(4-0)$

$= -6\hat{i} - 3\hat{j} + 4\hat{k}$

Step 3: Find Magnitude 1 Mark

$|\vec{AB} \times \vec{AC}| = \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$

Step 4: Calculate Area 1 Mark

Area of triangle $= \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{\sqrt{61}}{2}$ sq. units

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is:
(A) 0   (B) 1   (C) 3   (D) -1
Ans: (C) 3
Solution: $\hat{j} \times \hat{k} = \hat{i}$, so $\hat{i} \cdot \hat{i} = 1$. Similarly each term = 1. Total = 3
2022 (2 Marks): If $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$, prove that $\vec{a}$ and $\vec{b}$ are perpendicular.
Solution:
$|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$
$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$
$|\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{b}|^2$
$4\vec{a} \cdot \vec{b} = 0 \Rightarrow \vec{a} \cdot \vec{b} = 0$
Hence, $\vec{a} \perp \vec{b}$
2021 (3 Marks): Show that the points A(2, 3, -4), B(1, -2, 3), and C(3, 8, -11) are collinear.
Solution:
$\vec{AB} = -\hat{i} - 5\hat{j} + 7\hat{k}$
$\vec{AC} = \hat{i} + 5\hat{j} - 7\hat{k} = -\vec{AB}$
Since $\vec{AC} = -1 \times \vec{AB}$, vectors are collinear.
Hence A, B, C are collinear.
2020 (5 Marks): Find a unit vector perpendicular to each of the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Solution Steps:
1. Find $\vec{a} + \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$
2. Find $\vec{a} - \vec{b} = -\hat{j} - 2\hat{k}$
3. Calculate $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})$ using determinant
4. Find magnitude of the cross product
5. Divide by magnitude to get unit vector

Exam Strategy & Mistake Bank

Common Mistakes 🚨

Mistake 1: Confusing dot product with cross product. Dot product gives scalar, cross product gives vector!
Mistake 2: Wrong sign in cross product: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ (not commutative!)
Mistake 3: Forgetting to take $\frac{1}{2}$ in triangle area formula. Parallelogram = full, Triangle = half!
Mistake 4: Using external section formula when internal is required (or vice versa). Check the ratio carefully!
Mistake 5: Wrong determinant expansion in cross product. Always expand along first row!
Mistake 6: Not checking $l^2 + m^2 + n^2 = 1$ for direction cosines.

Scoring Tips 🏆

Tip 1: Always write formula first, then substitute values. Shows proper method!
Tip 2: For cross product, write the determinant clearly. Partial marks even if calculation wrong!
Tip 3: In angle problems, always mention $0 \leq \theta \leq \pi$ for dot product, $0 \leq \theta \leq \pi$ for cross.
Tip 4: Show unit vector calculation step: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$ explicitly.
Tip 5: For collinearity, show $\vec{AB} = k\vec{BC}$ or use cross product = 0.
Tip 6: Write final answer in simplest form with proper vector notation.

Formula Sheet (Must Remember!) 📝

Basic Formulas

1. Position Vector: $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$

2. Magnitude: $|\vec{r}| = \sqrt{x^2 + y^2 + z^2}$

3. Unit Vector: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$

4. Direction Cosines: $l = \frac{x}{r}, m = \ Chapter 10: Vector Algebra - Board Exam Notes | SJMaths
Home > Class 12 > Chapter 10: Vector Algebra
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Chapter 10: Vector Algebra

Complete Board Exam Focused Notes with All Formulas & PYQs

Exam Weightage & Blueprint

Total: ~14 Marks

Vector Algebra is a high-scoring chapter with direct formula-based questions. Master dot product, cross product, and section formula for guaranteed marks!

Question Type Marks Frequency Focus Topic
MCQ 1 Very High Magnitude, Unit Vector, Direction Cosines
Short Answer (2M) 2 Very High Dot Product, Position Vector, Section Formula
Short Answer (3M) 3 High Cross Product, Angle Between Vectors
Long Answer (5M) 5 Very High Area of Triangle/Parallelogram, Collinearity

⏰ Last 24-Hour Checklist

Basic Concepts (Must Know!)

  • ☐ Position vector: $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$
  • ☐ Magnitude: $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$
  • ☐ Unit vector: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$
  • ☐ Direction cosines: $l = \frac{x}{r}, m = \frac{y}{r}, n = \frac{z}{r}$
  • ☐ Property: $l^2 + m^2 + n^2 = 1$

Products (Very Important!)

  • ☐ Dot Product: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$
  • ☐ Component form: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$
  • ☐ Cross Product: $\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\, \hat{n}$
  • ☐ Section Formula (Internal): $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$
  • ☐ Area of triangle: $\frac{1}{2}|\vec{AB} \times \vec{AC}|$

Basic Concepts of Vectors ★★★★★

Definition: A quantity that has both magnitude and direction is called a vector.

Notation: $\vec{AB}$ or $\vec{a}$ (bold) or $\overrightarrow{AB}$

Types of Vectors

Zero Vector

$\vec{0}$ or $\vec{AA}$

Magnitude = 0, direction undefined

Unit Vector

$|\vec{a}| = 1$

$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$

Collinear Vectors

$\vec{b} = \lambda\vec{a}$

Parallel vectors

Position Vector & Components

Position Vector of P(x, y, z): $$\vec{OP} = \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$$ Magnitude: $$|\vec{r}| = \sqrt{x^2 + y^2 + z^2}$$
Unit Vectors along Axes:
$\hat{i}$ = unit vector along x-axis
$\hat{j}$ = unit vector along y-axis
$\hat{k}$ = unit vector along z-axis
Properties: $|\hat{i}| = |\hat{j}| = |\hat{k}| = 1$
$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$ (mutually perpendicular)

Direction Cosines & Direction Ratios

If $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $|\vec{r}| = r$, then:
Direction Cosines: $l = \cos\alpha = \frac{x}{r}$, $m = \cos\beta = \frac{y}{r}$, $n = \cos\gamma = \frac{z}{r}$
Important Property: $l^2 + m^2 + n^2 = 1$
Direction Ratios: $a = x, b = y, c = z$ (proportional to direction cosines)
Relationship: $$l = \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$

Vector Operations 🔥🔥🔥

Addition of Vectors

Triangle Law: If $\vec{AB}$ and $\vec{BC}$ are two vectors, then: $$\vec{AB} + \vec{BC} = \vec{AC}$$ Parallelogram Law: If $\vec{a}$ and $\vec{b}$ are adjacent sides of a parallelogram, their sum is the diagonal.
Properties of Vector Addition:
1. Commutative: $\vec{a} + \vec{b} = \vec{b} + \vec{a}$
2. Associative: $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$
3. Additive Identity: $\vec{a} + \vec{0} = \vec{a}$
4. Additive Inverse: $\vec{a} + (-\vec{a}) = \vec{0}$

Scalar Multiplication

If $\vec{a}$ is a vector and $\lambda$ is a scalar, then:
• $|\lambda\vec{a}| = |\lambda||\vec{a}|$
• Direction same as $\vec{a}$ if $\lambda > 0$, opposite if $\lambda < 0$
• $\lambda\vec{a} = \lambda(x\hat{i} + y\hat{j} + z\hat{k}) = \lambda x\hat{i} + \lambda y\hat{j} + \lambda z\hat{k}$

Component Form Operations

If $\vec{a