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.

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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.

Test yourself — a quick 12-question Vectors recall quiz, instant score.

Concept 1 of 4: Magnitude of a vector and distance between two points

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2019 · Q67Easy

Example 1 · Vectors · Magnitude, Components, Projection, and Direction Cosines

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}

$\overrightarrow{AB}$ 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

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2023 · Q67Easy

Example 2 · Vectors · Magnitude, Components, Projection, and Direction Cosines

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?

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 2 more on direction cosines

Concept 3 of 4: Scalar projection of one vector on another

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 $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}$ . When an unknown component is fixed by a stated projection, the SIGN is data: "the projection of $\vec{a}$ on $\vec{b}$ is $2$ " rejects any root of the resulting equation that makes $\vec{a}\cdot\vec{b}$ negative — squaring to solve introduces that spurious root, so substitute back before choosing.

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}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2023 · Q91Easy

Example 3 · Vectors · Magnitude, Components, Projection, and Direction Cosines

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}

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 2 more on scalar projection of one vector on another

Concept 4 of 4: Unit vectors and direction-given construction

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}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2021 · Q71Easy

Example 4 · Vectors · Magnitude, Components, Projection, and Direction Cosines

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?

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

Test yourself on Vectors

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Related notes

Cross product and triple product →