Chapter 10: Vector Algebra
Complete Board Exam Focused Notes with All Formulas & PYQs
Exam Weightage & Blueprint
Total: ~14 MarksVector 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)
$\hat{i}$ = 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)
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: $\vec{a} + (-\vec{a}) = \vec{0}$
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 $\lambda < 0$
• $\lambda\vec{a} = \lambda(x\hat{i} + y\hat{j} + z\hat{k}) = \lambda x\hat{i} + \lambda y\hat{j} + \lambda z\hat{k}$
• $|\lambda\vec{a}| = |\lambda||\vec{a}|$
• 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 $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ 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 $a_1 = b_1, a_2 = b_2, a_3 = b_3$
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!
• 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
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: $\vec{a} \parallel \vec{b} \iff \vec{a} = k\vec{b} \text{ or } \theta = 0 \text{ or } \pi$
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)
• 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!
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 $\vec{b}$: $\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
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
(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, $\vec{a} \perp \vec{b}$
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.
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
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: $l = \frac{x}{r}, m = \frac{y}{r}, n = \frac{z}{r}$ where $l^2 + m^2 + n^2 = 1$
5. Vector Joining Points: $\vec{P_1P_2} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$
Section Formula
6. Internal Division (m:n): $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$7. External Division (m:n): $\vec{r} = \frac{m\vec{b} - n\vec{a}}{m-n}$
8. Midpoint: $\vec{r} = \frac{\vec{a} + \vec{b}}{2}$
Dot Product
9. $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$10. $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$
11. Angle: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
12. Projection: $\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
13. Perpendicular: $\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$
Cross Product
14. $\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\, \hat{n}$15. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
16. Parallel: $\vec{a} \parallel \vec{b} \iff \vec{a} \times \vec{b} = \vec{0}$
17. Area of Triangle: $\frac{1}{2}|\vec{AB} \times \vec{AC}|$
18. Area of Parallelogram: $|\vec{a} \times \vec{b}|$
Unit Vector Relations
19. $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$20. $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$
Important Theorems & Results
Triangle Law of Vector Addition:
If two vectors are represented by two sides of a triangle taken in order, then their sum is represented by the third side taken in reverse order.
$\vec{AB} + \vec{BC} = \vec{AC}$
If two vectors are represented by two sides of a triangle taken in order, then their sum is represented by the third side taken in reverse order.
$\vec{AB} + \vec{BC} = \vec{AC}$
Parallelogram Law:
If two vectors are represented by two adjacent sides of a parallelogram, then their sum is represented by the diagonal passing through their common point.
If two vectors are represented by two adjacent sides of a parallelogram, then their sum is represented by the diagonal passing through their common point.
Cauchy-Schwarz Inequality:
$|\vec{a} \cdot \vec{b}| \leq |\vec{a}||\vec{b}|$
Triangle Inequality:
$|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$
Collinearity Condition:
Points A, B, C are collinear if and only if:
• $\vec{AB} = k\vec{BC}$ for some scalar k, OR
• $\vec{AB} \times \vec{AC} = \vec{0}$, OR
• $\vec{AB} + \vec{BC} = \vec{AC}$
Points A, B, C are collinear if and only if:
• $\vec{AB} = k\vec{BC}$ for some scalar k, OR
• $\vec{AB} \times \vec{AC} = \vec{0}$, OR
• $\vec{AB} + \vec{BC} = \vec{AC}$
Practice Problems (Self-Assessment)
Level 1: Basic (1-2 Marks Each)
Q1. Find the magnitude of $\vec{a} = 3\hat{i} - 2\hat{j} + 6\hat{k}$
Answer: $|\vec{a}| = \sqrt{9+4+36} = 7$
Q2. If $\vec{a} \cdot \vec{b} = 0$ and $|\vec{a}| = 3$, $|\vec{b}| = 4$, what is the angle between them?
Answer: $90°$ (perpendicular)
Q3. Find $\hat{i} \times (\hat{j} + \hat{k})$
Answer: $\hat{k} - \hat{j}$
Level 2: Intermediate (3-4 Marks Each)
Q4. Show that vectors $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}$ are parallel.
Hint: Show $\vec{b} = -2\vec{a}$
Q5. Find the position vector of point dividing join of A(1,2,3) and B(3,4,5) in ratio 2:1 internally.
Hint: Use section formula with m=2, n=1
Q6. If $|\vec{a}| = 3$, $|\vec{b}| = 4$, and $\vec{a} \cdot \vec{b} = 6$, find the angle between them.
Answer: $\theta = \cos^{-1}(1/2) = 60°$
Level 3: Advanced (5-6 Marks Each)
Q7. Find the area of parallelogram whose adjacent sides are $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$
Hint: Area = $|\vec{a} \times \vec{b}|$
Q8. Prove that if $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, then $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$
Hint: Express $\vec{c} = -(\vec{a} + \vec{b})$ and expand
Q9. Find a unit vector perpendicular to the plane containing vectors $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$
Hint: Find $\vec{a} \times \vec{b}$, then divide by its magnitude
Quick Revision Tips (1 Day Before Exam)
✓ Must Revise Topics
Formulas to Memorize:
- Magnitude formula
- Unit vector formula
- Direction cosines (3 formulas)
- Section formula (internal & external)
- Dot product (2 forms)
- Cross product (determinant form)
- Area formulas (triangle & parallelogram)
Common Question Types:
- Find unit vector (2M) ⭐⭐⭐
- Angle between vectors (3M) ⭐⭐⭐
- Prove perpendicular (3M) ⭐⭐⭐
- Area of triangle (5M) ⭐⭐⭐⭐⭐
- Collinearity proof (3M) ⭐⭐⭐⭐
- Section formula (2M) ⭐⭐⭐⭐
✓ Calculation Tips
- ✓ For cross product, always write determinant first - never skip this step!
- ✓ Check sign carefully when expanding determinant (remember −j term)
- ✓ For angle: First find dot product, then magnitudes, then apply formula
- ✓ Triangle area = $\frac{1}{2}$ × cross product magnitude (don't forget half!)
- ✓ Unit vector = divide by magnitude (show both numerator and denominator)
- ✓ Direction cosines: Always verify $l^2 + m^2 + n^2 = 1$ at the end
Time Management Per Question
1 Mark (MCQ) = 1-2 minutes2 Marks = 3-4 minutes
3 Marks = 5-6 minutes
5 Marks = 8-10 minutes
Total Vector Questions in Paper: Usually 3-4 questions = 10-14 marks
Exam Day Checklist 📋
Before Starting Vector Questions:
- □ Read question carefully - Is it dot or cross product?
- □ Check if unit vector or magnitude is asked
- □ For area questions, identify if triangle or parallelogram
- □ For section formula, check internal or external division
- □ Underline key words: perpendicular, parallel, collinear, angle
While Solving:
- □ Write formula first, then substitute values
- □ Show all calculation steps clearly
- □ For determinant, expand step-by-step
- □ Simplify radicals properly (√12 = 2√3)
- □ Write vector notation correctly with arrows or bold
- □ Box or underline final answer
Final Check:
- □ Did you write correct formula?
- □ Are calculations correct? (recheck once)
- □ For area, did you apply $\frac{1}{2}$ if triangle?
- □ Is answer in simplest form?
- □ Proper units mentioned if required?
- □ Vector notation used throughout?
Golden Rules for Full Marks 🏆
The 5 Commandments of Vector Algebra
1. Always Show the Formula FirstWrite the formula before substitution - gets you method marks even if calculation is wrong!
2. Master the Determinant
Cross product questions = easy 3-5 marks if you know determinant expansion. Practice 10 times!
3. Remember the "Half" Rule
Triangle area = HALF of cross product magnitude. Parallelogram = FULL. Don't mix!
4. Check Your Signs
In determinant expansion, middle term has MINUS sign. In $\vec{a} \times \vec{b}$, order matters!
5. Verify Direction Cosines
Always check $l^2 + m^2 + n^2 = 1$. Shows examiner you understand the concept!
Pro Tips from Toppers:
• "I always draw a small diagram for position vectors - helps visualize!" - Priya (98%)
• "Practice determinant expansion daily - it's free marks in exam!" - Rahul (95%)
• "Make a formula sheet and revise it every night" - Anjali (96%)
• "For angle questions, I always write both formulas (dot & cross) to choose" - Arjun (97%)
• "I always draw a small diagram for position vectors - helps visualize!" - Priya (98%)
• "Practice determinant expansion daily - it's free marks in exam!" - Rahul (95%)
• "Make a formula sheet and revise it every night" - Anjali (96%)
• "For angle questions, I always write both formulas (dot & cross) to choose" - Arjun (97%)