Vector Algebra Class 12: Notes, Methods and Solved Questions

What Vector Algebra is about

Weightage: This chapter is part of Unit IV, Vectors and Three-Dimensional Geometry, which carries 14 of the 80 theory marks together with Three Dimensional Geometry.Source: CBSE Mathematics syllabus 2026-27, Unit IV Vectors and Three-Dimensional Geometry

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7.1 marks per paper. See the questions

Key concepts

Vectors, scalars, magnitude and position vector

NCERT section 10.2

A vector is a directed line segment $\overrightarrow{AB}$ with initial point A and terminal point B; its length $|\overrightarrow{AB}|$ is the magnitude. Quantities with only magnitude are scalars.

The position vector of P(x, y, z) relative to the origin O is $\overrightarrow{OP}=\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, with magnitude

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

Worked example

Write the position vector of P(2, −3, 6) and find its magnitude. [1 mark]

Try it first, then show the solutionSolution

1. $\overrightarrow{OP}=2\hat{i}-3\hat{j}+6\hat{k}$
2. $|\overrightarrow{OP}|=\sqrt{4+9+36}=7$

Answer: $2\hat{i}-3\hat{j}+6\hat{k}$; magnitude 7

Direction cosines and direction ratios of a vector

NCERT section 10.2

If $\vec{r}=a\hat{i}+b\hat{j}+c\hat{k}$ makes angles α, β, γ with the positive x-, y- and z-axes, then its direction cosines are

$l=\cos \alpha =\frac{a}{|\vec{r}|},m=\cos \beta =\frac{b}{|\vec{r}|},n=\cos \gamma =\frac{c}{|\vec{r}|}$

and they always satisfy $l^2+m^2+n^2=1$. Any three numbers proportional to l, m, n (such as a, b, c themselves) are direction ratios. The unit vector along $\vec{r}$ is $l\hat{i}+m\hat{j}+n\hat{k}$.

Worked example

Find the direction cosines of $2\hat{i}-\hat{j}+2\hat{k}$. [1 mark]

Try it first, then show the solutionSolution

1. $|\vec{r}|=\sqrt{4+1+4}=3$
2. $l=\frac{2}{3}$, $m=-\frac{1}{3}$, $n=\frac{2}{3}$ (check: $\frac{4+1+4}{9}=1$)

Answer: $\frac{2}{3},-\frac{1}{3},\frac{2}{3}$

Types of vectors

NCERT section 10.3

• Zero vector $\vec{0}$: magnitude 0, no definite direction.
• Unit vector: magnitude 1; $\hat{a}=\frac{\vec{a}}{|\vec{a}|}$.
• Co-initial vectors: same initial point.
• Collinear (parallel) vectors: parallel to the same line, whatever their sizes or senses; $\vec{b}=\lambda \vec{a}$ for some scalar λ.
• Equal vectors: same magnitude and same direction (initial points may differ).
• Negative of a vector: same magnitude, opposite direction; $\overrightarrow{BA}=-\overrightarrow{AB}$.

In components, $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ are collinear exactly when $\frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3}$ (with zero components matching).

Worked example

Are $\vec{a}=2\hat{i}-4\hat{j}+6\hat{k}$ and $\vec{b}=-\hat{i}+2\hat{j}-3\hat{k}$ collinear? Are they equal? [1 mark]

Try it first, then show the solutionSolution

1. $\vec{b}=-\frac{1}{2}\vec{a}$, so they are collinear (opposite senses).
2. Their magnitudes differ (√56 and √14) and their directions are opposite, so they are not equal.

Answer: Collinear, not equal.

Addition of vectors and its properties

NCERT section 10.4

Triangle law: $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$. Parallelogram law: if $\vec{a}$ and $\vec{b}$ are adjacent sides from a common point, $\vec{a}+\vec{b}$ is the diagonal from that point. Around any closed triangle, $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}$.

Addition is commutative ($\vec{a}+\vec{b}=\vec{b}+\vec{a}$) and associative; $\vec{0}$ is the identity and $-\vec{a}$ the inverse. In components, add corresponding components.

For a parallelogram with adjacent sides $\vec{a},\vec{b}$, the diagonals are $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ (or $\vec{b}-\vec{a}$).

Worked example

If $\vec{a}=\hat{i}+2\hat{j}-\hat{k}$ and $\vec{b}=3\hat{i}-\hat{j}+2\hat{k}$, find the unit vector along $\vec{a}+\vec{b}$. [2 marks]

Try it first, then show the solutionSolution

1. $\vec{a}+\vec{b}=4\hat{i}+\hat{j}+\hat{k}$, magnitude $\sqrt{16+1+1}=3\sqrt{2}$

Answer: $\frac{1}{3\sqrt{2}}(4\hat{i}+\hat{j}+\hat{k})$

Multiplying by a scalar; vector of a given magnitude

NCERT section 10.5.1

$\lambda \vec{a}$ has magnitude $|\lambda ||\vec{a}|$, the same direction as $\vec{a}$ if λ > 0 and the opposite direction if λ < 0. In components, multiply each component by λ.

A vector of magnitude k in the direction of $\vec{a}$ is $k\hat{a}=\frac{k\vec{a}}{|\vec{a}|}$; opposite to $\vec{a}$ it is $-k\hat{a}$. The scalar components of $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ are x, y, z and the vector components are $x\hat{i},y\hat{j},z\hat{k}$.

Worked example

Find the vector of magnitude 9 in the direction of $2\hat{i}-\hat{j}+2\hat{k}$. [1 mark]

Try it first, then show the solutionSolution

1. Magnitude of the given vector is 3, so the unit vector is $\frac{1}{3}(2\hat{i}-\hat{j}+2\hat{k})$.
2. Multiply by 9: $6\hat{i}-3\hat{j}+6\hat{k}$

Answer: $6\hat{i}-3\hat{j}+6\hat{k}$

Vector joining two points

NCERT section 10.5.2

For $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$:

$\overrightarrow{P_1{P}_2}=\overrightarrow{O{P}_2}-\overrightarrow{O{P}_1}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

"Terminal minus initial". Its magnitude is the distance between the points.

Worked example

Find $\overrightarrow{PQ}$ and its magnitude for P(1, 2, −3) and Q(4, −2, 9). [1 mark]

Try it first, then show the solutionSolution

1. $\overrightarrow{PQ}=3\hat{i}-4\hat{j}+12\hat{k}$
2. $|\overrightarrow{PQ}|=\sqrt{9+16+144}=13$

Answer: $3\hat{i}-4\hat{j}+12\hat{k}$; 13 units

Section formula and collinearity

NCERT section 10.5.3

If A and B have position vectors $\vec{a}$ and $\vec{b}$, the point R dividing AB in the ratio m : n has position vector

• internally: $\vec{r}=\frac{m\vec{b}+n\vec{a}}{m+n}$
• externally: $\vec{r}=\frac{m\vec{b}-n\vec{a}}{m-n}$
• mid-point: $\frac{\vec{a}+\vec{b}}{2}$

Three points A, B, C are collinear if $\overrightarrow{AB}=\lambda \overrightarrow{AC}$ for some scalar λ.

Worked example

A and B have position vectors $\hat{i}+2\hat{j}-\hat{k}$ and $4\hat{i}-\hat{j}+5\hat{k}$. Find the position vectors of the points dividing AB in the ratio 2 : 1 (a) internally, (b) externally. [2 marks]

Try it first, then show the solutionSolution

1. (a) $\frac{2(4\hat{i}-\hat{j}+5\hat{k})+(\hat{i}+2\hat{j}-\hat{k})}{3}=\frac{9\hat{i}+9\hat{k}}{3}=3\hat{i}+3\hat{k}$
2. (b) $\frac{2(4\hat{i}-\hat{j}+5\hat{k})-(\hat{i}+2\hat{j}-\hat{k})}{2-1}=7\hat{i}-4\hat{j}+11\hat{k}$

Answer: (a) $3\hat{i}+3\hat{k}$ (b) $7\hat{i}-4\hat{j}+11\hat{k}$

Scalar (dot) product

NCERT section 10.6.1

$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$, where θ (0 ≤ θ ≤ π) is the angle between them. The result is a scalar.

• In components: $\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$
• $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$; $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$
• Commutative and distributive; $\vec{a}\cdot \vec{a}=|\vec{a}{|}^2$
• $\vec{a}\perp \vec{b}⟺\vec{a}\cdot \vec{b}=0$ (non-zero vectors)
• Angle: $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$; θ is acute if the dot product is positive, obtuse if negative.

Worked example

Find the angle between $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}+2\hat{k}$. [2 marks]

Try it first, then show the solutionSolution

1. $\vec{a}\cdot \vec{b}=2-1+2=3$; $|\vec{a}|=|\vec{b}|=\sqrt{6}$
2. $\cos \theta =\frac{3}{6}=\frac{1}{2}$

Answer: $\theta =\frac{\pi }{3}$

Projection of a vector on a line

NCERT section 10.6.2

The (scalar) projection of $\vec{a}$ on $\vec{b}$ is the length of the shadow of $\vec{a}$ on the line of $\vec{b}$:

$\text{projection of }\vec{a}\text{ on }\vec{b}=\vec{a}\cdot \hat{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$

The projection vector is $\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{b}{|}^2}\right)\vec{b}$. A negative projection means the angle is obtuse. Note that the projection of $\vec{a}$ on $\vec{b}$ is generally different from that of $\vec{b}$ on $\vec{a}$.

Worked example

Find the projection of $\vec{a}=2\hat{i}+3\hat{j}+2\hat{k}$ on $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$. [2 marks]

Try it first, then show the solutionSolution

1. $\vec{a}\cdot \vec{b}=2+6+2=10$; $|\vec{b}|=\sqrt{6}$
2. Projection $=\frac{10}{\sqrt{6}}=\frac{5\sqrt{6}}{3}$

Answer: $\frac{5\sqrt{6}}{3}$

Using the dot product with magnitudes

NCERT section 10.6.1

Expanding with the distributive law gives identities used in many short questions:

• $|\vec{a}+\vec{b}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2+2\vec{a}\cdot \vec{b}$
• $|\vec{a}-\vec{b}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2-2\vec{a}\cdot \vec{b}$
• $(\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=|\vec{a}{|}^2-|\vec{b}{|}^2$

Consequences: $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|⟺\vec{a}\perp \vec{b}$; if $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, then $\vec{c}=-(\vec{a}+\vec{b})$ and squaring gives the angle between $\vec{a}$ and $\vec{b}$.

Worked example

If $|\vec{a}|=2$, $|\vec{b}|=3$ and $\vec{a}\cdot \vec{b}=4$, find $|\vec{a}+\vec{b}|$ and $|\vec{a}-\vec{b}|$. [2 marks]

Try it first, then show the solutionSolution

1. $|\vec{a}+\vec{b}{|}^2=4+9+8=21$
2. $|\vec{a}-\vec{b}{|}^2=4+9-8=5$

Answer: $\sqrt{21}$ and $\sqrt{5}$

Vector (cross) product

NCERT section 10.6.3

$\vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta \hat{n}$, where $\hat{n}$ is the unit vector perpendicular to both, chosen by the right-hand rule. The result is a vector.

$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$

• $\hat{i}\times \hat{j}=\hat{k}$, $\hat{j}\times \hat{k}=\hat{i}$, $\hat{k}\times \hat{i}=\hat{j}$; $\hat{i}\times \hat{i}=\vec{0}$
• Not commutative: $\vec{b}\times \vec{a}=-\vec{a}\times \vec{b}$; distributive over addition
• $\vec{a}\parallel \vec{b}⟺\vec{a}\times \vec{b}=\vec{0}$
• Unit vectors perpendicular to both: $\pm \frac{\vec{a}\times \vec{b}}{|\vec{a}\times \vec{b}|}$; $\sin \theta =\frac{|\vec{a}\times \vec{b}|}{|\vec{a}||\vec{b}|}$

Worked example

Find $\vec{a}\times \vec{b}$ for $\vec{a}=\hat{i}+2\hat{j}-\hat{k}$, $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$, and the unit vectors perpendicular to both. [2 marks]

Try it first, then show the solutionSolution

1. $\vec{a}\times \vec{b}=\hat{i}(6-1)-\hat{j}(3+2)+\hat{k}(-1-4)=5\hat{i}-5\hat{j}-5\hat{k}$
2. Magnitude $5\sqrt{3}$; unit vectors $\pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})$

Answer: $5\hat{i}-5\hat{j}-5\hat{k}$; $\pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})$

Areas using the cross product

NCERT section 10.6.3

• Parallelogram with adjacent sides $\vec{a}$, $\vec{b}$: area $=|\vec{a}\times \vec{b}|$
• Parallelogram with diagonals $\overrightarrow{d_1}$, $\overrightarrow{d_2}$: area $=\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}|$
• Triangle ABC: area $=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

If the area of the triangle is 0, the three points are collinear.

Worked example

Find the area of the triangle with vertices A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1). [2 marks]

Try it first, then show the solutionSolution

1. $\overrightarrow{AB}=\hat{j}+2\hat{k}$, $\overrightarrow{AC}=\hat{i}+2\hat{j}$
2. $\overrightarrow{AB}\times \overrightarrow{AC}=\hat{i}(0-4)-\hat{j}(0-2)+\hat{k}(0-1)=-4\hat{i}+2\hat{j}-\hat{k}$, magnitude $\sqrt{21}$

Answer: $\frac{\sqrt{21}}{2}$ square units

Maths in real life: Vector Algebra

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Section Formula and Collinearity of Vectors
• Scalar (Dot) Product of Vectors
• Vector (Cross) Product of Vectors

Vector Algebra formula sheet

Magnitude

$|x\hat{i}+y\hat{j}+z\hat{k}|=\sqrt{x^2+y^2+z^2}$

Unit vector

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

Direction cosines

$l=\frac{a}{|\vec{r}|},m=\frac{b}{|\vec{r}|},n=\frac{c}{|\vec{r}|},l^2+m^2+n^2=1$

Vector joining two points

$\overrightarrow{P_1{P}_2}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

Section formula

$\text{internal: }\frac{m\vec{b}+n\vec{a}}{m+n},\text{external: }\frac{m\vec{b}-n\vec{a}}{m-n}$

Dot product

$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta =a_1{b}_1+a_2{b}_2+a_3{b}_3$

Angle between vectors

$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Projection

${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$

Cross product

$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|,|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$

Areas

$\text{parallelogram}=|\vec{a}\times \vec{b}|=\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}|,\text{triangle}=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

Perpendicular and parallel

$\vec{a}\perp \vec{b}⟺\vec{a}\cdot \vec{b}=0,\vec{a}\parallel \vec{b}⟺\vec{a}\times \vec{b}=\vec{0}$

Magnitude identities

$|\vec{a}\pm \vec{b}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2\pm 2\vec{a}\cdot \vec{b}$

How the board asks Vector Algebra

The question patterns to expect, and the method that works for each.

Tricks to remember Vector Algebra

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Terminal minus initial

  AB = (position vector of B) − (position vector of A). The arrow points to the first one you write.

  Use it when: Vector joining two points.

• Dot gives a number, cross gives an arrow

  a · b is a scalar (cos θ); a × b is a vector (sin θ) perpendicular to both.

  Use it when: Choosing the right product.

• i → j → k → i

  Going round the cycle gives +: i × j = k, j × k = i, k × i = j. Going backwards gives −.

  Use it when: Cross products of unit vectors.

• Zero tests

  Dot zero → perpendicular. Cross zero → parallel.

  Use it when: Perpendicularity and collinearity questions.

• Diagonals halve

  Area from diagonals is half the cross product; area of a triangle is half the cross product of two sides.

  Use it when: Area questions.

Vector Algebra: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Vector Algebra

The slips examiners see most often, and what to write instead.

• ✗Mistake: Reversing the vector between two points

  Computing A − B for AB.

  Wrong working

  A(2, −3, 5), B(3, −4, 7): $\overrightarrow{AB}=-\hat{i}+\hat{j}-2\hat{k}$

  Right working

  $\overrightarrow{AB}=\hat{i}-\hat{j}+2\hat{k}$

  ✓Do this: Always terminal minus initial.

  Usually costs: 1 mark, and every later step.

• ✗Mistake: Sign slip in the middle term of the cross product

  Forgetting the minus on the ĵ component of the determinant.

  ✓Do this: Write $\hat{i}(a_2{b}_3-a_3{b}_2)-\hat{j}(a_1{b}_3-a_3{b}_1)+\hat{k}(a_1{b}_2-a_2{b}_1)$ explicitly.

  Usually costs: 1-2 marks.

• ✗Mistake: Using the full cross product for area from diagonals

  Treating diagonals like adjacent sides.

  ✓Do this: Area = ½|d₁ × d₂| when diagonals are given.

  Usually costs: 1 mark.

• ✗Mistake: Giving only one unit vector perpendicular to both

  Forgetting the opposite direction.

  ✓Do this: Write ± in front; there are two such unit vectors.

  Usually costs: ½ to 1 mark.

• ✗Mistake: Mixing up projection of a on b and b on a

  Dividing by the wrong magnitude.

  ✓Do this: Projection of a on b divides by |b| (the vector you project onto).

  Usually costs: 1 mark.

NCERT exercise guide: Vector Algebra

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Vector Algebra: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7.1 marks per paper; 38 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Vector Algebra

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Vector Algebra question, by year