What is the weightage of Chapter 10 Vector Algebra in CBSE Class 12?

Vector Algebra and Chapter 11 3D Geometry together form Unit IV (14 marks). Chapter 10 accounts for 5 to 7 marks.

What is the geometric meaning of the dot product vs cross product?

Dot product (a . b) produces a scalar measuring alignment and projection (0 if vectors are orthogonal). Cross product (a x b) produces a vector perpendicular to both, whose magnitude equals the area of the spanned parallelogram.

How do you compute the projection of vector a on vector b?

Projection of a on b = (a . b) / |b|.

What is the area of a triangle formed by adjacent vector sides a and b?

Area of triangle = (1/2) * |a x b|.

Class 12 Maths Chapter 10: Vector Algebra Notes & Formulas

2026?27 Syllabus Focus

The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.

Vector magnitude and direction; direction ratios and cosines; vector types and position vectors.
Section formula and scalar (dot) product.
Vector (cross) product: geometric interpretation, properties and applications.

CBSE Mathematics (Code 041) curriculum, 2026?27

Last 24-Hour Checklist

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

$\vec{b} = λ\vec{a}$

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)

Direction Cosines & Direction Ratios

If $\vec{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ and $|\vec{r}| = r$, then:
Direction Cosines: $l = \cosα = \frac{x}{r}$, $m = \cosβ = \frac{y}{r}$, $n = \cosγ = \frac{z}{r}$
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 $\vec{BC}$ represent two consecutive sides of a triangle, then the third side taken in reverse direction represents their sum: $$\vec{AB} + \vec{BC} = \vec{AC}$$ Parallelogram Law: If two vectors $\vec{a}$ and $\vec{b}$ represent adjacent sides of a parallelogram, their sum $\vec{a} + \vec{b}$ is represented by the diagonal passing through their common initial point.

Triangle Law

$\vec{AC} = \vec{AB} + \vec{BC}$

Parallelogram Law

$Diagonal \vec{d} = \vec{a} + \vec{b}$

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 $λ$ is a scalar, then:
$|λ\vec{a}| = |λ||\vec{a}|$
Direction same as $\vec{a}$ if $λ > 0$, opposite if $λ < 0$
$λ\vec{a} = λ(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) = λ x\mathbf{i} + λ y\mathbf{j} + λ z\mathbf{k}$

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 $a_1 = b_1, a_2 = b_2, a_3 = b_3$

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!

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: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cosθ$ where $θ$ is the angle between $\vec{a}$ and $\vec{b}$, $0 \leq θ \leq π$

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	$\mathbf{i} · \mathbf{i} = \mathbf{j} · \mathbf{j} = \mathbf{k} · \mathbf{k} = 1$
Perpendicular	$\mathbf{i} · \mathbf{j} = \mathbf{j} · \mathbf{k} = \mathbf{k} · \mathbf{i} = 0$

Important Results

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

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

3. Parallel Vectors: $\vec{a} \parallel \vec{b} \iff \vec{a} = k\vec{b} or θ = 0 or π$

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

1. Scalar Projection of $\vec{a}$ on $\vec{b}$:

$p = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = |\vec{a}|\cosθ$

2. Vector Projection of $\vec{a}$ on $\vec{b}$:

$\vec{v} = ( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} ) \vec{b} = (Scalar Projection) \cdot \hat{b}$

⚡ High-Yield Vector Theorems & Board Identities:

Direction Cosines & Angles Identities (Top MCQ):
- Fundamental Identity: $l^2 + m^2 + n^2 = 1$ (where $l=\cos\alpha, m=\cos\beta, n=\cos\gamma$).
- Sine Identity: $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = (1-l^2) + (1-m^2) + (1-n^2) = 3 - 1 = \mathbf{2}$.
- Equally inclined vector to coordinate axes: $\cos\alpha = \cos\beta = \cos\gamma \implies 3l^2 = 1 \implies l = m = n = \pm \frac{1}{\sqrt{3}}$.
Projection of a Vector:
- Scalar projection of $\vec{a}$ on $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \vec{a} \cdot \hat{b}$.
- Vector projection of $\vec{a}$ along $\vec{b} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$.
Lagrange's Identity:
$$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$
Orthogonality & Collinearity Conditions:
- $\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$.
- $\vec{a} \parallel \vec{b} \iff \vec{a} \times \vec{b} = \vec{0} \iff \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$.
Sum Vector Magnitude & Angle Theorem:
- $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$
- If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, then: $$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{1}{2}(|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2)$$
Geometric Areas via Cross Product:
- Triangle with adjacent sides $\vec{a}, \vec{b}$: $\Delta = \frac{1}{2}|\vec{a} \times \vec{b}|$.
- Parallelogram with adjacent sides $\vec{a}, \vec{b}$: $\text{Area} = |\vec{a} \times \vec{b}|$.
- Parallelogram with diagonals $\vec{d_1}, \vec{d_2}$: $\text{Area} = \frac{1}{2}|\vec{d_1} \times \vec{d_2}|$.

Cross Product (Vector Product)

Definition: The cross product of two vectors $\vec{a}$ and $\vec{b}$ is: $\vec{a} × \vec{b} = |\vec{a}||\vec{b}|\sinθ\, \hat{n}$ where $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ (right-hand rule)

Component Form (Determinant Method): $\vec{a} × \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$ $= (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}$

Properties of Cross Product

Property	Formula
Not Commutative	$\vec{a} × \vec{b} = -(\vec{b} × \vec{a})$
Distributive	$\vec{a} × (\vec{b} + \vec{c}) = \vec{a} × \vec{b} + \vec{a} × \vec{c}$
Scalar Multiple	$k(\vec{a} × \vec{b}) = (k\vec{a}) × \vec{b} = \vec{a} × (k\vec{b})$
Self Cross	$\vec{a} × \vec{a} = \vec{0}$
Unit Vectors	$\mathbf{i} × \mathbf{j} = \mathbf{k}$, $\mathbf{j} × \mathbf{k} = \mathbf{i}$, $\mathbf{k} × \mathbf{i} = \mathbf{j}$
Reverse Order	$\mathbf{j} × \mathbf{i} = -\mathbf{k}$, $\mathbf{k} × \mathbf{j} = -\mathbf{i}$, $\mathbf{i} × \mathbf{k} = -\mathbf{j}$

Memory Trick for $\mathbf{i}, \mathbf{j}, \mathbf{k}$ Cross Products:
Cyclic order (clockwise): $\mathbf{i} × \mathbf{j} = \mathbf{k}$, $\mathbf{j} × \mathbf{k} = \mathbf{i}$, $\mathbf{k} × \mathbf{i} = \mathbf{j}$
Anti-cyclic (reverse): Just add negative sign!

Geometric Applications

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

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

3. Unit Vector Perpendicular to both $\vec{a}$ and $\vec{b}$: $\hat{n} = \frac{\vec{a} × \vec{b}}{|\vec{a} × \vec{b}|}$

Parallel Vectors: $\vec{a} \parallel \vec{b}$ if and only if $\vec{a} × \vec{b} = \vec{0}$

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

$\cosθ = \frac{\vec{a} · \vec{b}}{|\vec{a}||\vec{b}|} = \frac{1}{\sqrt{3} × \sqrt{3}} = \frac{1}{3}$

Step 4: Find Angle 0.5 Mark

$θ = \cos^{-1}(\frac{1}{3})$

Q3. Find $\vec{a} × \vec{b}$ 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

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

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

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

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

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

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

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

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

Step 3: Find Magnitude 1 Mark

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

Step 4: Calculate Area 1 Mark

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

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): The value of $\mathbf{i} · (\mathbf{j} × \mathbf{k}) + \mathbf{j} · (\mathbf{k} × \mathbf{i}) + \mathbf{k} · (\mathbf{i} × \mathbf{j})$ is:
(A) 0 (B) 1 (C) 3 (D) -1
Ans: (C) 3
Solution: $\mathbf{j} × \mathbf{k} = \mathbf{i}$, so $\mathbf{i} · \mathbf{i} = 1$. 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:
$|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$
$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$
$|\vec{a}|^2 + 2\vec{a} · \vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a} · \vec{b} + |\vec{b}|^2$
$4\vec{a} \cdot \vec{b} = 0 \Rightarrow \vec{a} \cdot \vec{b} = 0$
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 $\vec{AC} = -1 × \vec{AB}$, 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 $(\vec{a} + \vec{b}) × (\vec{a} - \vec{b})$ 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: $\vec{a} × \vec{b} = -(\vec{b} × \vec{a})$ (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 $0 \leq θ \leq π$ for dot product, $0 \leq θ \leq π$ 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. $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cosθ$

10. $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$

11. Angle: $\cosθ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

12. Projection: $proj_{\vec{b}}\vec{a} = \frac{\vec{a} · \vec{b}}{|\vec{b}|}$

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

Cross Product

14. $\vec{a} × \vec{b} = |\vec{a}||\vec{b}|\sinθ\, \hat{n}$

15. $\vec{a} × \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$

16. Parallel: $\vec{a} \parallel \vec{b} \iff \vec{a} × \vec{b} = \vec{0}$

17. Area of Triangle: $\frac{1}{2}|\vec{AB} × \vec{AC}|$

18. Area of Parallelogram: $|\vec{a} × \vec{b}|$

Unit Vector Relations

19. $\mathbf{i} · \mathbf{j} = \mathbf{j} · \mathbf{k} = \mathbf{k} · \mathbf{i} = 0$

20. $\mathbf{i} × \mathbf{j} = \mathbf{k}$, $\mathbf{j} × \mathbf{k} = \mathbf{i}$, $\mathbf{k} × \mathbf{i} = \mathbf{j}$

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
$\vec{AB} × \vec{AC} = \vec{0}$, 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: $90^°$ (perpendicular)

Q3. Find $\mathbf{i} × (\mathbf{j} + \mathbf{k})$

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: $θ = \cos^{-1}(1/2) = 60^°$

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 = $|\vec{a} × \vec{b}|$

Q8. Prove that if $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, then $\vec{a} × \vec{b} = \vec{b} × \vec{c} = \vec{c} × \vec{a}$

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 $\vec{a} × \vec{b}$, 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 minutes
2 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?

Chapter 10 Board Mastery Test

Practice test • 10 marks • 25 min

Q1 (1 Mark): Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.

Q2 (2 Marks): Find a unit vector perpendicular to both vectors $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} - 2\hat{k}$.

Q3 (3 Marks): If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$, find the angle between $\vec{a}$ and $\vec{b}$.

Q4 (4 Marks): The adjacent sides of a parallelogram are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Find its area and the unit vector parallel to its diagonal.

Marking Scheme & Solutions

Solution 1:

$Projection of \vec{a} on \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

$\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10$.

$|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$.

$Projection = \frac{10}{\sqrt{6}} = \mathbf{\frac{5\sqrt{6}}{3}}$. [1 Mark]

Solution 2:

$\vec{a} × \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} = \hat{i}(2 - 3) - \hat{j}(-8 + 6) + \hat{k}(4 - 2) = -\hat{i} + 2\hat{j} + 2\hat{k}$. [1 Mark]

$|\vec{a} × \vec{b}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{9} = 3$.

Unit vector $\hat{n} = \pm \frac{\vec{a} × \vec{b}}{|\vec{a} × \vec{b}|} = \mathbf{\pm \frac{1}{3}(-\hat{i} + 2\hat{j} + 2\hat{k})}$. [1 Mark]

Solution 3:

$\vec{a} + \vec{b} = -\vec{c} \implies |\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 = |\vec{c}|^2$. [1 Mark]

$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2 \implies 3^2 + 5^2 + 2|\vec{a}||\vec{b}|\cosθ = 7^2$.

$9 + 25 + 2(3)(5)\cosθ = 49 \implies 34 + 30\cosθ = 49 \implies 30\cosθ = 15$. [1 Mark]

$\cosθ = \frac{15}{30} = \frac{1}{2} \implies \mathbf{θ = \frac{π}{3} = 60^°}$. [1 Mark]

Solution 4:

$\vec{a} × \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1) = 5\hat{i} + \hat{j} - 4\hat{k}$. [1 Mark]

$Area = |\vec{a} × \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \mathbf{\sqrt{42} sq units}$. [1 Mark]

Diagonal $\vec{d} = \vec{a} + \vec{b} = (3+1)\hat{i} + (1-1)\hat{j} + (4+1)\hat{k} = 4\hat{i} + 5\hat{k}$. [1 Mark]

Unit vector in direction of diagonal $\hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{4\hat{i} + 5\hat{k}}{\sqrt{4^2 + 5^2}} = \mathbf{\frac{1}{\sqrt{41}}(4\hat{i} + 5\hat{k})}$. [1 Mark]

Chapter 10: Vector Algebra

Overview

2026?27 Syllabus Focus

Last 24-Hour Checklist

Basic Concepts (Must Know!)

Products (Very Important!)

Basic Concepts of Vectors

Types of Vectors

Zero Vector

Unit Vector

Collinear Vectors

Position Vector & Components

Direction Cosines & Direction Ratios

Vector Operations

Addition of Vectors

Triangle Law

Parallelogram Law

Scalar Multiplication

Component Form Operations

Vector Joining Two Points

Section Formula

Dot Product (Scalar Product)

Properties of Dot Product

Important Results

Projection of Vector

Cross Product (Vector Product)

Properties of Cross Product

Geometric Applications

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)

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)

Q3. Find $\vec{a} × \vec{b}$ 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)

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

Previous Year Questions (PYQs)

Exam Strategy & Mistake Bank

Common Mistakes

Scoring Tips

Formula Sheet (Must Remember!)

Basic Formulas

Section Formula

Dot Product

Cross Product

Unit Vector Relations

Important Theorems & Results

Practice Problems (Self-Assessment)

Level 1: Basic (1-2 Marks Each)

Level 2: Intermediate (3-4 Marks Each)

Level 3: Advanced (5-6 Marks Each)

Quick Revision Tips (1 Day Before Exam)

Must Revise Topics

Formulas to Memorize:

Common Question Types:

Time Management Per Question

Exam Day Checklist

Before Starting Vector Questions:

While Solving:

Final Check:

Chapter 10 Board Mastery Test

Marking Scheme & Solutions

Chapter 10 Study Ecosystem: Complete Preparation Hub

Frequently Asked Questions: Class 12 Vector Algebra