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!)
Products (Very Important!)
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):
$$\vec{OP} = \vec{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$$
Magnitude:
$$|\vec{r}| = \sqrt{x^2 + y^2 + z^2}$$
Unit Vectors along Axes:
$\mathbf{i}$ = unit vector along x-axis
$\mathbf{j}$ = unit vector along y-axis
$\mathbf{k}$ = unit vector along z-axis
Properties: $|\mathbf{i}| = |\mathbf{j}| = |\mathbf{k}| = 1$
$\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0$ (mutually perpendicular)
$\mathbf{i}$ = unit vector along x-axis
$\mathbf{j}$ = unit vector along y-axis
$\mathbf{k}$ = unit vector along z-axis
Properties: $|\mathbf{i}| = |\mathbf{j}| = |\mathbf{k}| = 1$
$\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0$ (mutually perpendicular)
Direction Cosines & Direction Ratios
If $\vec{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ 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 $\lambda < 0$
$\lambda\vec{a} = \lambda(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) = \lambda x\mathbf{i} + \lambda y\mathbf{j} + \lambda z\mathbf{k}$
$|\lambda\vec{a}| = |\lambda||\vec{a}|$
Direction same as $\vec{a}$ if $\lambda > 0$, opposite if $\lambda < 0$
$\lambda\vec{a} = \lambda(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) = \lambda x\mathbf{i} + \lambda y\mathbf{j} + \lambda z\mathbf{k}$
Component Form Operations
If $\vec{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\vec{b} = b_1\mathbf{i} +
b_2\mathbf{j} +
b_3\mathbf{k}$, then:
Addition: $\vec{a} + \vec{b} = (a_1+b_1)\mathbf{i} + (a_2+b_2)\mathbf{j} + (a_3+b_3)\mathbf{k}$
Subtraction: $\vec{a} - \vec{b} = (a_1-b_1)\mathbf{i} + (a_2-b_2)\mathbf{j} + (a_3-b_3)\mathbf{k}$
Equal Vectors: $\vec{a} = \vec{b}$ if and only if $a_1 = b_1, a_2 = b_2, a_3 = b_3$
Addition: $\vec{a} + \vec{b} = (a_1+b_1)\mathbf{i} + (a_2+b_2)\mathbf{j} + (a_3+b_3)\mathbf{k}$
Subtraction: $\vec{a} - \vec{b} = (a_1-b_1)\mathbf{i} + (a_2-b_2)\mathbf{j} + (a_3-b_3)\mathbf{k}$
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 $P_2(x_2, y_2, z_2)$:
$\vec{P_1P_2} = (x_2-x_1)\mathbf{i} + (y_2-y_1)\mathbf{j} + (z_2-z_1)\mathbf{k}$
$|\vec{P_1P_2}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$
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 | $\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1$ |
| Perpendicular | $\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0$ |
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):
$\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3
\\ b_1 &
b_2 & b_3 \end{vmatrix}$
$= (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}$
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 | $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, $\mathbf{k} \times \mathbf{i} = \mathbf{j}$ |
| Reverse Order | $\mathbf{j} \times \mathbf{i} = -\mathbf{k}$, $\mathbf{k} \times \mathbf{j} = -\mathbf{i}$, $\mathbf{i} \times \mathbf{k} = -\mathbf{j}$ |
Memory Trick for $\mathbf{i}, \mathbf{j}, \mathbf{k}$ Cross Products:
Cyclic order (clockwise): $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, $\mathbf{k} \times \mathbf{i} = \mathbf{j}$
Anti-cyclic (reverse): Just add negative sign!
Cyclic order (clockwise): $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, $\mathbf{k} \times \mathbf{i} = \mathbf{j}$
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 $\vec{a} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$ (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
$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\mathbf{i} + 3\mathbf{j} + \mathbf{k}}{\sqrt{14}}$
$= \frac{2}{\sqrt{14}}\mathbf{i} + \frac{3}{\sqrt{14}}\mathbf{j} + \frac{1}{\sqrt{14}}\mathbf{k}$
Q2. Find the angle between vectors $\vec{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}$ and $\vec{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}$ (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 $\vec{a} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ and $\vec{b} = 3\mathbf{i} + 5\mathbf{j} - 2\mathbf{k}$ (3 Marks)
Solution: 3 Marks
$\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix}$
$= \mathbf{i}(1 \times (-2) - 3 \times 5) - \mathbf{j}(2 \times (-2) - 3 \times 3) + \mathbf{k}(2 \times 5 - 1 \times 3)$
$= \mathbf{i}(-2 - 15) - \mathbf{j}(-4 - 9) + \mathbf{k}(10 - 3)$
$= -17\mathbf{i} + 13\mathbf{j} + 7\mathbf{k}$
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
$\vec{AB} = (2-1)\mathbf{i} + (3-1)\mathbf{j} + (5-2)\mathbf{k} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$
$\vec{AC} = (1-1)\mathbf{i} + (5-1)\mathbf{j} + (5-2)\mathbf{k} = 4\mathbf{j} + 3\mathbf{k}$
Step 2: Calculate Cross Product 2 Marks
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix}$
$= \mathbf{i}(6-12) - \mathbf{j}(3-0) + \mathbf{k}(4-0)$
$= -6\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$
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 $\mathbf{i} \cdot (\mathbf{j} \times \mathbf{k}) +
\mathbf{j}
\cdot (\mathbf{k} \times \mathbf{i}) + \mathbf{k} \cdot (\mathbf{i} \times \mathbf{j})$ is:
(A) 0 (B) 1 (C) 3 (D) -1
Ans: (C) 3
Solution: $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, so $\mathbf{i} \cdot \mathbf{i} = 1$. Similarly each term = 1. Total = 3
(A) 0 (B) 1 (C) 3 (D) -1
Ans: (C) 3
Solution: $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, so $\mathbf{i} \cdot \mathbf{i} = 1$. 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:
$\vec{AB} = -\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}$
$\vec{AC} = \mathbf{i} + 5\mathbf{j} - 7\mathbf{k} = -\vec{AB}$
Since $\vec{AC} = -1 \times \vec{AB}$, vectors are collinear.
Hence A, B, C are collinear.
Solution:
$\vec{AB} = -\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}$
$\vec{AC} = \mathbf{i} + 5\mathbf{j} - 7\mathbf{k} = -\vec{AB}$
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 $\vec{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}$ and
$\vec{b} =
\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$.
Solution Steps:
1. Find $\vec{a} + \vec{b} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$
2. Find $\vec{a} - \vec{b} = -\mathbf{j} - 2\mathbf{k}$
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 $\vec{a} + \vec{b} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$
2. Find $\vec{a} - \vec{b} = -\mathbf{j} - 2\mathbf{k}$
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\mathbf{i} + y\mathbf{j} + z\mathbf{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: $\vec{P_1P_2} = (x_2-x_1)\mathbf{i} + (y_2-y_1)\mathbf{j} + (z_2-z_1)\mathbf{k}$
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. $\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
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. $\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0$20. $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, $\mathbf{k} \times \mathbf{i} = \mathbf{j}$
Important Theorems & Results
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 $\vec{a} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}$
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: $90^\circ$ (perpendicular)
Q3. Find $\mathbf{i} \times (\mathbf{j} + \mathbf{k})$
Answer: $\mathbf{k} - \mathbf{j}$
Level 2: Intermediate (3-4 Marks Each)
Q4. Show that vectors $\vec{a} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ and $\vec{b} = -4\mathbf{i} + 6\mathbf{j} - 8\mathbf{k}$ 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: $\theta = \cos^{-1}(1/2) = 60^\circ$
Level 3: Advanced (5-6 Marks Each)
Q7. Find the area of parallelogram whose adjacent sides are $\vec{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$ and $\vec{b} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$
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 $\vec{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$ and $\vec{b} = \mathbf{i} - \mathbf{j} + 2\mathbf{k}$
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)
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?