Chapter 10 · Vector Algebra · Go for it

Vectors, the quick way

A vector is a length with a direction, written as $x\hat i + y\hat j + z\hat k$ ⁠. Learn ten moves and you can answer most questions in this chapter.

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The chapter in one minute

What a vector is, and what the exam asks.

A vector is a length with a direction, like "40 km towards the north-east". A plain number with no direction, like 10 kg, is a scalar.

Every vector in this chapter is written with three parts, as $x\hat i + y\hat j + z\hat k$ ⁠. Once a vector is in this form, every question is just arithmetic.

There are two ways to multiply vectors. The dot product gives a number, and it finds angles and projections. The cross product gives a vector, and it finds areas and perpendicular directions.

Most exam questions on this chapter are one of ten kinds, and each kind has a fixed recipe. You will learn all ten below.

10question types

13formula cards

0is the dot product of two perpendicular vectors

1determinant does every cross product

Formulas

The formulas you must know

Learn these cards. Every question in the chapter is built from them. Here $\vec a = a_1\hat i + a_2\hat j + a_3\hat k$ and $\vec b = b_1\hat i + b_2\hat j + b_3\hat k$ ⁠.

Magnitude (length)

|\vec a| = \sqrt{a_1^2 + a_2^2 + a_3^2}

Square each part, add, then take the square root.

Unit vector

\hat a = \frac{1}{|\vec a|}\,\vec a

Divide a nonzero vector by its length. The result has length 1 and the same direction.

From A to B

\begin{aligned} \overrightarrow{AB} &= (x_2 - x_1)\hat i \\ &\quad + (y_2 - y_1)\hat j \\ &\quad + (z_2 - z_1)\hat k \end{aligned}

For $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ ⁠. End minus start: $\overrightarrow{AB} = \vec b - \vec a$ ⁠.

Direction cosines

l = \frac{a}{r},\ m = \frac{b}{r},\ n = \frac{c}{r}

For a vector with parts $a, b, c$ (its direction ratios) and length $r$ ⁠. Check: $l^2 + m^2 + n^2 = 1$ ⁠.

Section formula, inside

\vec r = \frac{m\vec b + n\vec a}{m + n}

R divides PQ in the ratio $m : n$ ⁠, where $\vec a$ and $\vec b$ are the position vectors of P and Q. The midpoint is $\frac{\vec a + \vec b}{2}$ ⁠.

Section formula, outside

\vec r = \frac{m\vec b - n\vec a}{m - n}

The same, with two minus signs. Here $m \ne n$ ⁠.

Dot product

\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta

The answer is a number. $\theta$ is the angle between the vectors.

Dot product in parts

\begin{aligned} &\vec a\cdot\vec b \\ &= a_1b_1 + a_2b_2 + a_3b_3 \end{aligned}

Multiply matching parts and add. Also $\vec a\cdot\vec a = |\vec a|^2$ ⁠.

Angle between two vectors

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

If two nonzero vectors have dot product 0, the angle is $\frac{\pi}{2}$ ⁠: they are perpendicular.

Projection of

\vec a

\vec b

\vec a\cdot\hat b = \frac{\vec a\cdot\vec b}{|\vec b|}

Divide by the length of the vector you project on.

Cross product

\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}

The answer is a vector, perpendicular to both. Order matters: $\vec b\times\vec a = -\vec a\times\vec b$ ⁠.

Size of the cross product

|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta

This is the area of the parallelogram with adjacent sides $\vec a$ and $\vec b$ ⁠.

Area of a triangle

\text{Area} = \frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|

Half of the parallelogram on two of its sides.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. A unit vector, or a vector of a given length

Spot it: "unit vector in the direction of", or "a vector of magnitude 8 in the direction of".

Find the length $|\vec a|$ with the magnitude card.
Divide the vector by its length. That is the unit vector $\hat a$ ⁠.
For a length of 8, multiply the unit vector by 8.

Exercise 10.2 Q10. Find a vector in the direction of $5\hat i - \hat j + 2\hat k$ which has magnitude 8 units.

Let $\vec a = 5\hat i - \hat j + 2\hat k$ ⁠.

\begin{aligned} |\vec a| &= \sqrt{5^2 + 1^2 + 2^2} \\ &= \sqrt{30} \end{aligned}

\hat a = \frac{1}{\sqrt{30}}(5\hat i - \hat j + 2\hat k)

\begin{aligned} 8\hat a &= \frac{8}{\sqrt{30}}(5\hat i - \hat j + 2\hat k) \\ &= \frac{40}{\sqrt{30}}\hat i - \frac{8}{\sqrt{30}}\hat j \\ &\quad + \frac{16}{\sqrt{30}}\hat k \end{aligned}

∴ $\frac{40}{\sqrt{30}}\hat i - \frac{8}{\sqrt{30}}\hat j + \frac{16}{\sqrt{30}}\hat k$

See a full exam answer · Go deeper: see it visually · Basics: length by Pythagoras

In past papers (5): click to see

Board 2026 Q23 · Board 2025 Q25(b) · Board 2023 Q22 · Board 2022 T2 Q7(b) · Sample 2020-21 Q10

2. The vector from A to B, and its direction cosines

Spot it: two points are given, like $A(1, 2, -3)$ and $B(-1, -2, 1)$ ⁠.

Find $\overrightarrow{AB}$ ⁠: end minus start, part by part.
Find its length with the magnitude card.
Direction cosines: divide each part by the length.

Exercise 10.2 Q13. Find the direction cosines of the vector joining the points $A(1, 2, -3)$ and $B(-1, -2, 1)$ ⁠, directed from A to B.

\begin{aligned} \overrightarrow{AB} &= (-1 - 1)\hat i + (-2 - 2)\hat j \\ &\quad + (1 + 3)\hat k \\ &= -2\hat i - 4\hat j + 4\hat k \end{aligned}

|\overrightarrow{AB}| = \sqrt{4 + 16 + 16} = 6

l, m, n = \frac{-2}{6}, \frac{-4}{6}, \frac{4}{6}

∴ $l, m, n = -\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}$

See a full exam answer · Go deeper: see it visually · Basics: points in space

In past papers (6): click to see

Board 2026 Q23 · Board 2025 Q25(b) · Board 2024 Q15 · Board 2023 Q23(a) · Sample 2026-27 Q14 · Sample 2024-25 Q24(b)

3. Section formula: the point that divides PQ

Spot it: "divides the line joining P and Q in the ratio 2 : 1", internally or externally.

Name the position vectors: $\vec a$ for P, $\vec b$ for Q. Write $m$ and $n$ from the ratio.
Inside: use the plus card. Outside: use the minus card.
Put the vectors in, collect $\hat i$ ⁠, $\hat j$ ⁠, $\hat k$ ⁠, and divide.

Exercise 10.2 Q15. P and Q have position vectors $\hat i + 2\hat j - \hat k$ and $-\hat i + \hat j + \hat k$ ⁠. Find the position vector of R, which divides PQ in the ratio 2 : 1 (i) internally, (ii) externally.

Here $\vec a = \hat i + 2\hat j - \hat k$ ⁠, $\vec b = -\hat i + \hat j + \hat k$ ⁠, $m = 2$ and $n = 1$ ⁠.

\begin{aligned} \vec r &= \frac{m\vec b + n\vec a}{m + n} \\ &= \frac{2(-\hat i + \hat j + \hat k)}{3} \\ &\quad + \frac{\hat i + 2\hat j - \hat k}{3} \\ &= \frac{-\hat i + 4\hat j + \hat k}{3} \end{aligned}

\begin{aligned} \vec r &= \frac{m\vec b - n\vec a}{m - n} \\ &= 2(-\hat i + \hat j + \hat k) \\ &\quad - (\hat i + 2\hat j - \hat k) \end{aligned}

∴ (i) $\vec r = -\frac{1}{3}\hat i + \frac{4}{3}\hat j + \frac{1}{3}\hat k$ ⁠, (ii) $\vec r = -3\hat i + 3\hat k$

See a full exam answer · Go deeper: see it visually · Basics: dividing in a ratio

In past papers (2): click to see

Board 2025 Q29(b) · Sample 2025-26 Q25

4. The angle between two vectors

Spot it: "find the angle between the vectors".

Work out the dot product $\vec a\cdot\vec b$ ⁠, part by part.
Work out both lengths.
Use the angle card. If $\cos\theta$ is not a standard value, leave the answer as $\cos^{-1}$ of it.

Exercise 10.3 Q2. Find the angle between the vectors $\hat i - 2\hat j + 3\hat k$ and $3\hat i - 2\hat j + \hat k$ ⁠.

Let $\vec a = \hat i - 2\hat j + 3\hat k$ and $\vec b = 3\hat i - 2\hat j + \hat k$ ⁠.

\begin{aligned} \vec a\cdot\vec b &= 1\cdot 3 + (-2)(-2) + 3\cdot 1 \\ &= 10 \end{aligned}

|\vec a| = \sqrt{1 + 4 + 9} = \sqrt{14}

|\vec b| = \sqrt{9 + 4 + 1} = \sqrt{14}

\begin{aligned} \cos\theta &= \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} \\ &= \frac{10}{14} = \frac{5}{7} \end{aligned}

∴ $\theta = \cos^{-1}\frac{5}{7}$

See a full exam answer · Go deeper: see it visually · Basics: cos as a shadow

In past papers (5): click to see

Board 2025 Q11 · Board 2025 Q25(a) · Sample 2024-25 Q9 · Sample 2024-25 Q28(a) · Sample 2020-21 Q12

5. Perpendicular vectors: set the dot product to 0

Spot it: "is perpendicular to", with an unknown such as $\lambda$ to find.

Write the vector with $\lambda$ in it, part by part.
Its dot product with the other vector is 0. Write that line.
Solve the equation for $\lambda$ ⁠.

Exercise 10.3 Q10. If $\vec a = 2\hat i + 2\hat j + 3\hat k$ ⁠, $\vec b = -\hat i + 2\hat j + \hat k$ and $\vec c = 3\hat i + \hat j$ are such that $\vec a + \lambda\vec b$ is perpendicular to $\vec c$ ⁠, find $\lambda$ ⁠.

\begin{aligned} \vec a + \lambda\vec b &= (2 - \lambda)\hat i + (2 + 2\lambda)\hat j \\ &\quad + (3 + \lambda)\hat k \end{aligned}

It is perpendicular to $\vec c$ ⁠, and $\vec c$ has no $\hat k$ part, so

\begin{aligned} &(\vec a + \lambda\vec b)\cdot\vec c = 0 \\ &\Rightarrow 3(2 - \lambda) + 1(2 + 2\lambda) = 0 \\ &\Rightarrow 8 - \lambda = 0 \end{aligned}

∴ $\lambda = 8$

See a full exam answer · Go deeper: see it visually

In past papers (6): click to see

Board 2026 Q13 · Board 2025 Q2 · Board 2023 Q8 · Sample 2024-25 Q24(a) · Sample 2023-24 Q16 · Sample 2022-23 Q23(a)

6. Projection: the shadow of one vector on another

Spot it: "find the projection of the vector $\vec a$ on the vector $\vec b$ ".

Work out the dot product $\vec a\cdot\vec b$ ⁠.
Work out the length of $\vec b$ ⁠, the vector you project on.
Divide: $\text{projection} = \frac{\vec a\cdot\vec b}{|\vec b|}$ ⁠.

Exercise 10.3 Q4. Find the projection of the vector $\hat i + 3\hat j + 7\hat k$ on the vector $7\hat i - \hat j + 8\hat k$ ⁠.

Let $\vec a = \hat i + 3\hat j + 7\hat k$ and $\vec b = 7\hat i - \hat j + 8\hat k$ ⁠.

\vec a\cdot\vec b = 7 - 3 + 56 = 60

|\vec b| = \sqrt{49 + 1 + 64} = \sqrt{114}

\begin{aligned} \text{projection} &= \frac{\vec a\cdot\vec b}{|\vec b|} \\ &= \frac{60}{\sqrt{114}} \end{aligned}

∴ $\text{projection} = \frac{60}{\sqrt{114}}$

See a full exam answer · Go deeper: see it visually · Basics: cos as a shadow

In past papers (4): click to see

Sample 2025-26 Q13 · Sample 2024-25 Q28(a) · Sample 2023-24 Q12 · Sample 2022-23 Q8

7. Dot products with letters: expand like brackets

Spot it: no $\hat i, \hat j, \hat k$ at all, just $\vec a$ ⁠, $\vec b$ ⁠, their lengths and a dot product.

Expand the brackets as in algebra. The order does not matter: $\vec a\cdot\vec b = \vec b\cdot\vec a$ ⁠.
Replace every $\vec a\cdot\vec a$ by $|\vec a|^2$ ⁠.
Put in the given numbers and solve.

Exercise 10.3 Q9. Find $|\vec x|$ ⁠, if for a unit vector $\vec a$ ⁠, $(\vec x - \vec a)\cdot(\vec x + \vec a) = 12$ ⁠.

Expand. The middle terms cancel, because $\vec x\cdot\vec a = \vec a\cdot\vec x$ ⁠.

\begin{aligned} &(\vec x - \vec a)\cdot(\vec x + \vec a) \\ &= \vec x\cdot\vec x - \vec a\cdot\vec a \\ &= |\vec x|^2 - |\vec a|^2 \end{aligned}

$\vec a$ is a unit vector, so $|\vec a| = 1$ ⁠, and

\begin{aligned} &|\vec x|^2 - 1 = 12 \\ &\Rightarrow |\vec x|^2 = 13 \end{aligned}

∴ $|\vec x| = \sqrt{13}$

See a full exam answer · Go deeper: see it visually · Basics: brackets with dot products

In past papers (8): click to see

Board 2025 Q11 · Board 2024 Q24 · Board 2022 T2 Q7(a) · Sample 2024-25 Q10 · Sample 2022-23 Q17 · Sample 2022-23 Q25 · Sample 2021-22 T2 Q3 · Sample 2020-21 Q12

8. Cross product, and the area of a parallelogram

Spot it: "find $\vec a\times\vec b$ ", "find $|\vec a\times\vec b|$ ", or "the area of the parallelogram whose adjacent sides are".

Write the determinant: $\hat i, \hat j, \hat k$ on top, then the parts of $\vec a$ ⁠, then the parts of $\vec b$ ⁠.
Expand along the top row. The middle term has a minus: $\hat i(\ldots) - \hat j(\ldots) + \hat k(\ldots)$ ⁠.
For the area, find the length of the answer.

Exercise 10.4 Q10. Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec a = \hat i - \hat j + 3\hat k$ and $\vec b = 2\hat i - 7\hat j + \hat k$ ⁠.

\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} \\ &= \hat i(-1 + 21) - \hat j(1 - 6) \\ &\quad + \hat k(-7 + 2) \\ &= 20\hat i + 5\hat j - 5\hat k \end{aligned}

\begin{aligned} |\vec a\times\vec b| &= \sqrt{400 + 25 + 25} \\ &= \sqrt{450} = 15\sqrt{2} \end{aligned}

∴ $\text{Area} = 15\sqrt{2}$ sq. units

See a full exam answer · Go deeper: see it visually · Basics: determinants

In past papers (6): click to see

Board 2025 Q23 · Board 2024 Q25 · Board 2022 T2 Q7(b) · Sample 2025-26 Q25 · Sample 2024-25 Q25 · Sample 2020-21 Q26

9. Area of a triangle: half the cross product

Spot it: "the area of the triangle with vertices A, B and C".

Find two sides from the same corner: $\overrightarrow{AB}$ and $\overrightarrow{AC}$ ⁠.
Find $\overrightarrow{AB}\times\overrightarrow{AC}$ and its length.
Halve it, and write sq. units.

Exercise 10.4 Q9. Find the area of the triangle with vertices $A(1, 1, 2)$ ⁠, $B(2, 3, 5)$ and $C(1, 5, 5)$ ⁠.

\begin{aligned} \overrightarrow{AB} &= \hat i + 2\hat j + 3\hat k \\ \overrightarrow{AC} &= 4\hat j + 3\hat k \end{aligned}

\begin{aligned} &\overrightarrow{AB}\times\overrightarrow{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) \\ &\quad + \hat k(4 - 0) \\ &= -6\hat i - 3\hat j + 4\hat k \end{aligned}

\begin{aligned} |\overrightarrow{AB}\times\overrightarrow{AC}| &= \sqrt{36 + 9 + 16} \\ &= \sqrt{61} \end{aligned}

∴ $\text{Area} = \frac{1}{2}\sqrt{61}$ sq. units

See a full exam answer · Book Example 24 · Go deeper: see it visually

In past papers (2): click to see

Sample 2022-23 Q3 · Sample 2020-21 Q11

10. A unit vector perpendicular to two vectors

Spot it: "a unit vector perpendicular to each of the vectors".

Their cross product $\vec c$ is perpendicular to both.
Find its length $|\vec c|$ ⁠.
Divide, and put $\pm$ in front: there are two directions, up and down.

Exercise 10.4 Q2. Find a unit vector perpendicular to each of the vectors $\vec a + \vec b$ and $\vec a - \vec b$ ⁠, where $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$ ⁠.

\begin{aligned} \vec a + \vec b &= 4\hat i + 4\hat j \\ \vec a - \vec b &= 2\hat i + 4\hat k \end{aligned}

\begin{aligned} \vec c &= (\vec a + \vec b)\times(\vec a - \vec b) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} \\ &= \hat i(16 - 0) - \hat j(16 - 0) \\ &\quad + \hat k(0 - 8) \\ &= 16\hat i - 16\hat j - 8\hat k \end{aligned}

\begin{aligned} |\vec c| &= \sqrt{256 + 256 + 64} \\ &= \sqrt{576} = 24 \end{aligned}

∴ $\pm\frac{\vec c}{|\vec c|} = \pm\left(\frac{2}{3}\hat i - \frac{2}{3}\hat j - \frac{1}{3}\hat k\right)$

See a full exam answer · Book Example 23 · Go deeper: see it visually

In past papers (1): click to see

Sample 2026-27 Q15

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Write every vector with all three parts first. A missing part is 0: $4\hat j + 3\hat k = 0\hat i + 4\hat j + 3\hat k$ ⁠.
Add, subtract or multiply by a number part by part: $2(\hat i - \hat j) + 3\hat j = 2\hat i + \hat j$ ⁠. Two vectors are equal only when all three parts match.
Check a unit vector: its parts squared must add up to 1, for example $\frac{1}{9} + \frac{4}{9} + \frac{4}{9} = 1$ ⁠.
Check a cross product: dot it with each of the two vectors. Both answers must be 0.
Parallel (collinear) vectors: one is a number times the other, like $-4\hat i + 6\hat j - 8\hat k = -2(2\hat i - 3\hat j + 4\hat k)$ ⁠.
Brackets with dot products expand as in algebra: $(\vec a + \vec b)\cdot(\vec a - \vec b) = |\vec a|^2 - |\vec b|^2$ ⁠.
Remember $\hat i\cdot\hat i = 1$ and $\hat i\cdot\hat j = 0$ ⁠, and go round the circle $\hat i \to \hat j \to \hat k$ ⁠: $\hat i\times\hat j = \hat k$ ⁠, $\hat j\times\hat k = \hat i$ ⁠, $\hat k\times\hat i = \hat j$ ⁠. Against the circle, add a minus.
Three points collinear? Show that $\overrightarrow{AB}$ is a number times $\overrightarrow{BC}$ ⁠. Right-angled triangle? Find the side whose dot product with another side is 0.
No calculators: leave surds like $\sqrt{14}$ and angles like $\cos^{-1}\frac{5}{7}$ as they are.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Taking $\overrightarrow{AB}$ as start minus end → it is end minus start: $\overrightarrow{AB} = \vec b - \vec a$ ⁠.
Adding the parts to get the length → square, add, then take the square root.
Forgetting the minus on the $\hat j$ term of the determinant → $\hat i(\ldots) - \hat j(\ldots) + \hat k(\ldots)$ ⁠.
Swapping the order of a cross product → it changes the sign: $\vec b\times\vec a = -\vec a\times\vec b$ ⁠.
Dividing by the wrong length in a projection → on $\vec b$ ⁠, divide by $|\vec b|$ ⁠.
Giving $|\overrightarrow{AB}\times\overrightarrow{AC}|$ as the area of a triangle → halve it.
Giving only one perpendicular unit vector → write $\pm$ ⁠: both directions count.
Swapping $m$ and $n$ in the section formula → for $PR : RQ = m : n$ ⁠, $m$ goes with $\vec b$ ⁠, the position vector of Q.

Marks

How the examiner gives marks

From CBSE's own marking schemes and rules. Write these lines and the marks follow.

Vectors and three-dimensional geometry together carry 14 of the 80 marks in the paper.
Marks go to the lines of working, not just the answer. Write the formula you use, then the line with the numbers in it: the dot product, each length, the determinant.
"Show that" questions (collinear points, perpendicular vectors, a right-angled triangle) end with the conclusion in words, such as "Hence the points are collinear." That line carries ½ mark.
Give the final answer on its own line: a vector as $x\hat i + y\hat j + z\hat k$ ⁠, an angle in exact form, an area with sq. units.
A slip is penalised only once, so keep going. Any correct method gets full marks.
An MCQ earns 1 mark for the right option only; the working earns nothing.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 10.2 Q1 magnitudes of three vectors
Exercise 10.2 Q12 direction cosines of a vector
Exercise 10.2 Q8 the vector from P to Q, then its unit vector
Exercise 10.2 Q16 the midpoint, by the section formula
Exercise 10.3 Q1 an angle from the dot product and the lengths
Exercise 10.3 Q3 a projection
Exercise 10.3 Q5 unit vectors that are perpendicular: dot products are 0
Miscellaneous Q6 a vector of length 5 parallel to a sum
Exercise 10.3 Q6 expanding brackets with dot products
Exercise 10.4 Q1 the size of a cross product
Example 24 the area of a triangle from its corners
Miscellaneous Q12 a vector perpendicular to two vectors, with one more condition (the book's answer should have $-70\hat k$ ⁠)

You can do this

Ten moves, one at a time. Work through the practice list, check each answer with the tips above, and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Past papers

48 CBSE board and sample-paper questions on Vector Algebra, each with a full-marks exam answer. Practise them →