NDA Maths · Vectors

Magnitude, Components, Projection, Direction Cosines

How to measure a vector's length, decompose it along axes, project it onto another vector, and read off the angles it makes with the coordinate axes.

All Vectors notes NDA Maths strategy

Why this matters

This is where you learn to MEASURE vectors. How long is one (its magnitude)? At what angles to the coordinate axes does it point (its direction cosines)? How much of one vector lies along another (its scalar projection)? And how do you build a unit vector pointing exactly where you want it? Each of these turns the geometric arrow from the foundations into a number you can compute with. 11 PYQs across 2018–2024, almost entirely EASY or MODERATE — the formulas are short and the trap surface is narrow, so it's also the lowest-hanging-fruit subtopic in the chapter.

Concept 1 of 4

Magnitude of a vector and distance between two points

Intuition

The magnitude of a vector is its length — extended Pythagoras applied to its components. The distance between two points is the magnitude of the displacement vector joining them, which is also why

\overrightarrow{AB}=\vec{b}-\vec{a}

: subtract the tail's position vector from the head's.

Definition

If $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$ , then $|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ . For two points $A, B$ with position vectors $\vec{a}, \vec{b}$ : $\overrightarrow{AB} = \vec{b} - \vec{a}$ and $AB = |\vec{b} - \vec{a}|$ .

Magnitude and distance

|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \qquad AB = |\vec{b} - \vec{a}|

$v_1, v_2, v_3$ components of $\vec{v}$ along $\hat{i}, \hat{j}, \hat{k}$
$\vec{a}, \vec{b}$ position vectors of the endpoints

Diagram · magnitude = √(x² + y²)

The components x and y are the legs of a right triangle; the vector is the hypotenuse, so |v| = √(x² + y²) = √(16 + 9) = 5. In 3-D the same idea adds a third leg: |v| = √(x² + y² + z²).

Worked example

Position vectors of points

A

and

B

are

\vec{a} = 2\hat{i} + \hat{j} - \hat{k}

and

\vec{b} = 5\hat{i} + 5\hat{j} + 11\hat{k}

. Find the length

AB

Subtract position vectors: $\overrightarrow{AB} = \vec{b} - \vec{a} = (5-2)\hat{i} + (5-1)\hat{j} + (11-(-1))\hat{k} = 3\hat{i} + 4\hat{j} + 12\hat{k}$ .
Square each component: $3^2 + 4^2 + 12^2 = 9 + 16 + 144 = 169$ .
Take the square root: $AB = \sqrt{169} = 13$ .

Answer:

AB = 13

units

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the magnitude of

\vec{v} = 2\hat{i} - 3\hat{j} + 6\hat{k}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$|\vec{v}|$ for $\vec{v} = 3\hat{i} + 4\hat{j}$ ?
2.
$|\vec{v}|$ for $\vec{v} = \hat{i} + 2\hat{j} + 2\hat{k}$ ?
3.
Distance $AB$ for $A(0,0,0)$ , $B(1,2,2)$ ?
4.
$|\vec{v}|$ for $\vec{v} = 6\hat{i} + 8\hat{j}$ ?

From the bank · past-year question

Example 1VectorsEASY

If the position vectors of points A and B are

3\hat{i}-2\hat{j}+\hat{k}

and

2\hat{i}+4\hat{j}-3\hat{k}

respectively, then what is the length of

\overrightarrow{AB}

[Q67 · Apr · 2019]

$\overrightarrow{AB}$ is $\vec{b} - \vec{a}$ — head minus tail

Reverse the subtraction and you get

\overrightarrow{BA}

— the magnitude is the same but the displacement points the other way. Direction matters whenever the result feeds into a dot product or angle.

Lagrange identity gives you the missing magnitude

Whenever a question gives

|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = k

together with one magnitude, use the identity

|\vec{a}|^2|\vec{b}|^2 = |\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2

to read the other magnitude off directly. (The same identity is the central formula of

\vec{a}\times\vec{b}

magnitude work — see the cross-product note.)

Drill 3 more on magnitude of a vector and distance between two points

Concept 2 of 4

Direction Cosines

Intuition

If a vector makes angles

\alpha, \beta, \gamma

with the positive

x, y, z

axes, its direction cosines are

\cos\alpha, \cos\beta, \cos\gamma

— exactly the components of the unit vector along

\vec{v}

. Their squares sum to 1 (Pythagoras on the unit sphere). From this, the sum of the sines squared is forced to be 2 — the most-tested sister identity.

Definition

For $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$ of magnitude $|\vec{v}|$ , the direction cosines are $l = \cos\alpha = \dfrac{v_1}{|\vec{v}|}, \; m = \cos\beta = \dfrac{v_2}{|\vec{v}|}, \; n = \cos\gamma = \dfrac{v_3}{|\vec{v}|}$ . They satisfy $l^2 + m^2 + n^2 = 1$ , and consequently $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$ .

Direction-cosine identities

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \qquad \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2

$\alpha, \beta, \gamma$ angles between $\vec{v}$ and the positive $x, y, z$ axes
$l, m, n$ direction cosines (the unit vector's components)

Diagram · direction cosines (drag to rotate)

α ≈ 49° · l = 0.66β ≈ 62° · m = 0.48γ ≈ 54° · n = 0.58

l, m, n are the cosines of the angles r makes with the x-, y-, z-axes — and the components of the unit vector along r. So l² + m² + n² = 1.00 = 1, always.

Worked example

A vector

\vec{a} = 4\hat{i} - 8\hat{j} + \hat{k}

makes angles

\alpha, \beta, \gamma

with the positive axes. Find

\cos\alpha

and verify the identity

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

Magnitude: $|\vec{a}| = \sqrt{16 + 64 + 1} = \sqrt{81} = 9$ .
Direction cosines: $\cos\alpha = \tfrac{4}{9}$ , $\cos\beta = -\tfrac{8}{9}$ , $\cos\gamma = \tfrac{1}{9}$ .
Sum of squares: $\dfrac{16}{81} + \dfrac{64}{81} + \dfrac{1}{81} = \dfrac{81}{81} = 1$ . $\checkmark$

Answer:

\cos\alpha = \dfrac{4}{9}

; identity holds.

Practice this conceptself-check · 4 quick reps

Try it yourself

A vector makes

60^\circ

with the

x

-axis and

60^\circ

with the

y

-axis. Find

\cos\gamma

, where

\gamma

is the angle with the

z

-axis.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\cos\alpha$ for $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ ?
2.
What does $l^2 + m^2 + n^2$ equal?
3.
If $l = \tfrac{1}{3}$ , $m = \tfrac{2}{3}$ , find $n$ (positive).
4.
$\sin^2\alpha + \sin^2\beta + \sin^2\gamma = ?$

From the bank · past-year question

Example 2VectorsEASY

Consider the following for the items that follow: Let a vector

\vec{a}=4\hat{i}-8\hat{j}+\hat{k}

make angles

\alpha

\beta

\gamma

with the positive directions of

x

y

z

axes respectively.

What is

\cos\alpha

equal to?

[Q67 · Apr · 2023]

Factor-of-2 trap: $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$ , not 1

From

\sum\cos^2 = 1

and

\sin^2 = 1 - \cos^2

, summing three times:

\sum\sin^2 = 3 - \sum\cos^2 = 3 - 1 = 2

. The distractor

= 1

(copying the cosine identity) is the single most common wrong answer in this concept.

Direction cosines can be negative

An obtuse angle with an axis gives a negative cosine — totally fine. Some students try to force

l, m, n \geq 0

; don't. The identity

l^2 + m^2 + n^2 = 1

holds with signs.

Drill 1 more on direction cosines

Concept 3 of 4

Scalar projection of one vector on another

Intuition

The scalar projection of

\vec{a}

\vec{b}

measures how far

\vec{a}

reaches in the direction of

\vec{b}

— drop a perpendicular from the tip of

\vec{a}

onto the line through

\vec{b}

, and the signed length from the foot back to the origin is the projection. Algebraically it's just

(\vec{a}\cdot\vec{b})/|\vec{b}|

, where the dot product

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

is the sum of products of corresponding components (it gets its own full treatment in the next subtopic) — divided by the length you're projecting onto.

Definition

The scalar projection of $\vec{a}$ on $\vec{b}$ is $\text{proj}_{\vec{b}}\vec{a} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ , with $\vec{a}\cdot\vec{b} = a_1b_1 + a_2b_2 + a_3b_3$ . It is a signed scalar (positive when the projection lands in the direction of $\vec{b}$ , negative when opposite). The corresponding vector projection — projecting and keeping a vector — is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}$ .

Scalar and vector projection

\text{proj}_{\vec{b}}\vec{a} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|} \qquad \overrightarrow{\text{proj}_{\vec{b}}\vec{a}} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\,\vec{b}

$\vec{a}$ vector being projected
$\vec{b}$ vector providing the direction
$|\vec{b}|$ magnitude of $\vec{b}$ (NOT $|\vec{b}|^2$ for scalar version)

Worked example

Find the scalar projection of

\vec{a} = 3\hat{i} + \hat{j} - \hat{k}

\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}

Dot product: $\vec{a}\cdot\vec{b} = 3\cdot 2 + 1\cdot 2 + (-1)\cdot 1 = 6 + 2 - 1 = 7$ .
Magnitude of $\vec{b}$ : $|\vec{b}| = \sqrt{4 + 4 + 1} = 3$ .
Scalar projection: $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \dfrac{7}{3}$ .

Answer:

\dfrac{7}{3}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the scalar projection of

\vec{a} = \hat{i} + \hat{j} + \hat{k}

\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Scalar projection of $\vec{a}$ on $\vec{b}$ — formula?
2.
$\vec{a}\cdot\vec{b} = 6$ , $|\vec{b}| = 3$ . Scalar projection?
3.
Projection of $\vec{a} = 2\hat{i} + \hat{j}$ on $\vec{b} = \hat{i}$ ?
4.
If the scalar projection is negative, the angle is?

From the bank · past-year question

Example 3VectorsEASY

What is the length of projection of the vector

\hat{i}+2\hat{j}+3\hat{k}

on the vector

2\hat{i}+3\hat{j}-2\hat{k}

[Q91 · Sep · 2023]

Divide by $|\vec{b}|$ , not $|\vec{b}|^2$ , for the scalar projection

Vector projection has

|\vec{b}|^2

in the denominator because it carries the direction

\vec{b}

back into the answer; scalar projection drops the direction and divides only once. Mixing the two is a frequent factor-of-

|\vec{b}|

bug.

Sign of the scalar projection encodes obtuse/acute

If the projection comes out negative, the angle between

\vec{a}

and

\vec{b}

is obtuse. Don't reach for absolute value automatically — the sign is the information.

Drill 1 more on scalar projection of one vector on another

Concept 4 of 4

Unit vectors and direction-given construction

Intuition

To strip a vector of its length and keep only its direction, divide by its magnitude — the result is a unit vector. To go the other way, multiply a unit vector by the desired magnitude. When a question describes a vector by its angles with axes, use the cosines of those angles as the components of its unit form, then scale.

Definition

For any non-zero $\vec{v}$ , the unit vector along $\vec{v}$ is $\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}$ . A vector of magnitude $r$ making angles $\alpha, \beta, \gamma$ with the positive axes is $\vec{v} = r(\cos\alpha\,\hat{i} + \cos\beta\,\hat{j} + \cos\gamma\,\hat{k})$ . Special 2-D case: a unit vector in the $xy$ -plane at angle $\theta$ to the $x$ -axis is $\cos\theta\,\hat{i} + \sin\theta\,\hat{j}$ .

Unit vector and direction construction

\hat{v} = \dfrac{\vec{v}}{|\vec{v}|} \qquad \vec{v} = r(\cos\alpha\,\hat{i} + \cos\beta\,\hat{j} + \cos\gamma\,\hat{k})

$\hat{v}$ unit vector along $\vec{v}$
$r$ desired magnitude of the constructed vector
$\alpha, \beta, \gamma$ angles with the positive coordinate axes

Worked example

Find a vector of magnitude 15 in the direction of

4\hat{i} - 3\hat{k}

Magnitude of the direction vector: $|\vec{v}| = \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5$ .
Unit vector: $\hat{v} = \dfrac{4\hat{i} - 3\hat{k}}{5}$ .
Scale to magnitude 15: $15\,\hat{v} = 3(4\hat{i} - 3\hat{k}) = 12\hat{i} - 9\hat{k}$ .

Answer:

12\hat{i} - 9\hat{k}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the unit vector along

\vec{v} = 3\hat{i} - 4\hat{j}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Unit vector along $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ ?
2.
Vector of magnitude $10$ along $3\hat{i} + 4\hat{j}$ ?
3.
Unit vector in the $xy$ -plane at angle $\theta$ to the $x$ -axis?
4.
Unit vector along $5\hat{i}$ ?

From the bank · past-year question

Example 4VectorsEASY

A vector

\vec{r} = a\hat{i} + b\hat{j}

is equally inclined to both

x

and

y

axes. If the magnitude of the vector is 2 units, then what are the values of

a

and

b

respectively?

[Q71 · Apr · 2021]

Check that the given angles are consistent with $\sum\cos^2 = 1$

A vector cannot make

\alpha = 60^\circ

and

\beta = 45^\circ

with the

x

and

y

axes AND have

\gamma

acute unless the third cosine fits. From

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

\cos^2\gamma = 1 - \tfrac{1}{4} - \tfrac{1}{2} = \tfrac{1}{4}

, so

\gamma = 60^\circ

(acute) or

120^\circ

Equally inclined to two axes only fixes one component pair

\"Vector inclined equally to

x

and

y

axes\" means

a_1 = a_2

, which combined with the magnitude pins down both. Don't read it as

a_1 = a_2 = a_3

— that's three axes, a different constraint.

Drill 2 more on unit vectors and direction-given construction

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Magnitude of a vector and distance between two points
Magnitude and distance
$|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \qquad AB = |\vec{b} - \vec{a}|$
Direction Cosines
Direction-cosine identities
$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \qquad \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$
Scalar projection of one vector on another
Scalar and vector projection
$\text{proj}_{\vec{b}}\vec{a} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|} \qquad \overrightarrow{\text{proj}_{\vec{b}}\vec{a}} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\,\vec{b}$
Unit vectors and direction-given construction
Unit vector and direction construction
$\hat{v} = \dfrac{\vec{v}}{|\vec{v}|} \qquad \vec{v} = r(\cos\alpha\,\hat{i} + \cos\beta\,\hat{j} + \cos\gamma\,\hat{k})$

Watch out for (8)

$\overrightarrow{AB}$ is $\vec{b} - \vec{a}$ — head minus tail
→ Magnitude of a vector and distance between two points
Lagrange identity gives you the missing magnitude
→ Magnitude of a vector and distance between two points
Factor-of-2 trap: $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$ , not 1
→ Direction Cosines
Direction cosines can be negative
→ Direction Cosines
Divide by $|\vec{b}|$ , not $|\vec{b}|^2$ , for the scalar projection
→ Scalar projection of one vector on another
Sign of the scalar projection encodes obtuse/acute
→ Scalar projection of one vector on another
Check that the given angles are consistent with $\sum\cos^2 = 1$
→ Unit vectors and direction-given construction
Equally inclined to two axes only fixes one component pair
→ Unit vectors and direction-given construction

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1VectorsMODERATE

If the position vectors of

A

and

B

are

(\sqrt{2}-1)\hat{i}-\hat{j}

and

\hat{i}+(\sqrt{2}+1)\hat{j}

respectively, then what is the magnitude of

\overrightarrow{AB}

[Q70 · Sep · 2021]

Example 2VectorsMODERATE

Consider the following for the items that follow: Let a vector

\vec{a}=4\hat{i}-8\hat{j}+\hat{k}

make angles

\alpha

\beta

\gamma

with the positive directions of

x

y

z

axes respectively.

What is

\cos2\beta+\cos2\gamma

equal to?

[Q68 · Apr · 2023]

Example 3VectorsEASY

What is the scalar projection of

\vec{a} = \hat{i} - 2\hat{j} + \hat{k}

\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}

[Q46 · Sep · 2019]

Example 4VectorsHARD

If a vector of magnitude 2 units makes an angle

\dfrac{\pi}{3}

with

2\hat{i}

\dfrac{\pi}{4}

with

3\hat{j}

and an acute angle

\theta

with

4\hat{k}

, then what are the components of the vector?

[Q67 · Apr · 2024]

Example 5VectorsMODERATE

|\vec{a}|=3, |\vec{b}|=4

and

|\vec{a}-\vec{b}|=5

, then what is the value of

|\vec{a}+\vec{b}|

[Q66 · Sep · 2018]

Drill every past-year question on this subtopic

11 questions from the bank — paginated, with cart and Word-export support.

Drill the 11 questions

Drill every past-year question on this subtopic

Related notes