MHT-CET Maths · Vectors

Magnitude, Components, and Unit Vectors

How to write a vector in î ĵ k̂ component form, measure its length, build a unit vector pointing in any direction, and — the MHT-CET workhorse — find the magnitude of a sum of vectors from given angles or perpendicularity conditions.

All Vectors notes MHT-CET Maths strategy

Why this matters

This is the on-ramp to the whole Vectors chapter: every later technique (dot product, cross product, scalar triple product) starts by writing vectors in component form and reading off a magnitude. Across the 9 PYQs here, ONE shape dominates — finding the length of a combination like a + b + c by expanding |a + b + c|² = Σ|·|² + 2Σ(a·b) and using the angle or perpendicularity data to evaluate the dot-product cross terms. Master that single identity (plus the unit-vector-along-a-diagonal construction) and you have the entire subtopic; the difficulty mix is MODERATE-to-HARD, but it's the same expansion every time.

Concept 1 of 6

Vectors and component form

Intuition

Every vector in 3-D space can be built from three standard perpendicular unit vectors

\hat{i}, \hat{j}, \hat{k}

(pointing along the

x, y, z

axes). Writing a vector as

a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

is just saying "go

a_1

steps along

x

a_2

along

y

a_3

along

z

Definition

A vector $\vec{a}$ is written in component form as $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ , where:

$a_1, a_2, a_3$ are its components (also called scalar components) along the axes,
$\hat{i}, \hat{j}, \hat{k}$ are the standard basis — mutually perpendicular unit vectors of length 1.

A null (zero) vector $\vec{0}$ has all components zero and no direction; a unit vector has magnitude 1; collinear (parallel) vectors are scalar multiples of each other.

Component form

\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

$a_1, a_2, a_3$ components along $\hat{i}, \hat{j}, \hat{k}$
$\hat{i}, \hat{j}, \hat{k}$ standard perpendicular unit vectors

Diagram · component form (drag to rotate)

Step along x (2.4 î), then y ( 1.6 ĵ), then z ( 2.0 k̂) to reach the tip: v = 2.4 î + 1.6 ĵ + 2.0 k̂. Any vector is the sum of its axis components, and |v| = √(x² + y² + z²).

Worked example

Write the vector from the origin to the point

P(-1, 5, 2)

in component form, and state its components.

The position vector of $P$ goes $-1$ along $x$ , $5$ along $y$ , $2$ along $z$ .
In component form: $\overrightarrow{OP} = -\hat{i} + 5\hat{j} + 2\hat{k}$ .
Components: $a_1 = -1$ , $a_2 = 5$ , $a_3 = 2$ .

Answer:

\overrightarrow{OP} = -\hat{i} + 5\hat{j} + 2\hat{k}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
Component form of the vector to $P(3, 0, -4)$ ?
2.
What is the magnitude of each of $\hat{i}, \hat{j}, \hat{k}$ ?
3.
Are $2\hat{i} + 4\hat{j}$ and $\hat{i} + 2\hat{j}$ parallel?
4.
Components of $\hat{j} - \hat{k}$ ?

A missing axis means a zero component, not a 2-D vector

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

lives in 3-D with

a_2 = 0

. When you square components for a magnitude, the missing term contributes

0^2 = 0

— don't drop the slot or miscount the axes.

Concept 2 of 6

Magnitude of a vector and distance between two points

Intuition

The magnitude of a vector is its length — extended Pythagoras on its components. The displacement from

A

B

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

(head minus tail), and the distance

AB

is the magnitude of that displacement.

Definition

For $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ , the magnitude is $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$ . For points $A, B$ with position vectors $\vec{a}, \vec{b}$ : the displacement is $\overrightarrow{AB} = \vec{b} - \vec{a}$ and the distance is $AB = |\vec{b} - \vec{a}|$ .

Magnitude and distance

|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \qquad AB = |\vec{b} - \vec{a}|

$a_1, a_2, a_3$ components of $\vec{a}$
$\vec{a}, \vec{b}$ position vectors of $A$ and $B$

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

Points

A

and

B

have position vectors

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

and

\vec{b} = 5\hat{i} + 5\hat{j} + 14\hat{k}

. Find the distance

AB

Displacement: $\overrightarrow{AB} = \vec{b} - \vec{a} = 4\hat{i} + 3\hat{j} + 12\hat{k}$ .
Square the components: $4^2 + 3^2 + 12^2 = 16 + 9 + 144 = 169$ .
Distance: $AB = \sqrt{169} = 13$ .

Answer:

AB = 13

units

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the magnitude of

\vec{v} = 6\hat{i} - 2\hat{j} + 3\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 $A(0,0,0)$ , $B(2,3,6)$ ?
4.
$|\vec{v}|$ for $\vec{v} = 5\hat{i} + 12\hat{j}$ ?

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

Reversing it gives

\overrightarrow{BA}

. The distance (magnitude) is the same either way, but the direction flips — and direction matters the moment the result feeds a dot product or an angle.

Magnitude needs every component squared

For

3\hat{i} + 4\hat{j}

, the answer is

\sqrt{9 + 16} = 5

, not

3 + 4 = 7

. Add the SQUARES, then take ONE square root at the end.

Concept 3 of 6

Unit vector along a given direction

Intuition

To keep a vector's direction but throw away its length, divide it by its own magnitude — the result is a unit vector (length 1). To go the other way, multiply a unit vector by whatever magnitude you want.

Definition

For any non-zero $\vec{a}$ , the unit vector along it is $\hat{a} = \dfrac{\vec{a}}{|\vec{a}|}$ . A vector of magnitude $r$ in the same direction as $\vec{a}$ is $r\,\hat{a} = \dfrac{r}{|\vec{a}|}\,\vec{a}$ . Every unit vector's components square-sum to 1.

Unit vector

\hat{a} = \dfrac{\vec{a}}{|\vec{a}|} \qquad r\,\hat{a} = \dfrac{r}{|\vec{a}|}\,\vec{a}

$\hat{a}$ unit vector along $\vec{a}$
$|\vec{a}|$ magnitude of $\vec{a}$
$r$ desired magnitude of the scaled vector

Visualization · slide k, scale the vector

k = 2.0

k·v = (4.0, 2.0)|k·v| = |k|·|v| = 4.47

Multiplying by k scales the length by |k| and keeps the same line. k > 1 stretches, 0 < k < 1 shrinks, k < 0 flips to the opposite direction, and k = 0 collapses it to the zero vector.

Worked example

Find a vector of magnitude 10 in the direction of

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

Magnitude: $|\vec{a}| = \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5$ .
Unit vector: $\hat{a} = \dfrac{1}{5}(4\hat{i} - 3\hat{j})$ .
Scale to magnitude 10: $10\,\hat{a} = 2(4\hat{i} - 3\hat{j}) = 8\hat{i} - 6\hat{j}$ .

Answer:

8\hat{i} - 6\hat{j}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the unit vector along

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

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 $15$ along $3\hat{i} + 4\hat{j}$ ?
3.
Unit vector along $6\hat{k}$ ?
4.
What does the sum of squares of a unit vector's components equal?

Divide by the magnitude, don't subtract it

The unit vector is

\vec{a}/|\vec{a}|

— scale every component by the same

1/|\vec{a}|

. A common slip is normalising only one component or dividing by the wrong length.

Concept 4 of 6

Magnitude of a sum from angles or perpendicularity

Intuition

To find the length of a combination like

\vec{a} + \vec{b} + \vec{c}

, never compute components — instead SQUARE the magnitude. The square expands into the sum of each vector's squared length plus twice every pairwise dot product. Here we treat

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

as a recalled Std-11 fact (the dot product gets its full treatment on a later page); all you need is that perpendicular vectors have dot product 0, and that a given angle fixes each cross term.

Definition

Expand the square of the magnitude:

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

Each cross term is

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

. Two ways the cross terms get pinned down:

Given an angle $\theta$ between each pair → each dot product is $|\cdot||\cdot|\cos\theta$ .
Perpendicularity conditions like $\vec{a}\perp(\vec{b}+\vec{c})$ , $\vec{b}\perp(\vec{c}+\vec{a})$ , $\vec{c}\perp(\vec{a}+\vec{b})$ → adding the three gives $2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$ , so ALL cross terms vanish together and $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2$ .

Magnitude of a sum (squared)

|\vec{a} + \vec{b} + \vec{c}|^2 = \sum|\vec{a}|^2 + 2\sum(\vec{a}\cdot\vec{b}), \qquad \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta

$\sum|\vec{a}|^2$ $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2$
$\sum(\vec{a}\cdot\vec{b})$ all three pairwise dot products
$\theta$ angle between the pair of vectors in a dot product

Visualization · add two vectors tip-to-tail

a₁: 4a₂: 1b₁: 1b₂: 3

a + b = (5, 4)|a + b| = 6.40|a| + |b| = 7.29

The dashed amber arrow is b moved to the head of a — the resultant (indigo) runs from the shared tail to that head, which is also the diagonal of the parallelogram. Notice |a + b| equals |a| + |b| only when a and b point the same way.

Worked example

|\vec{a}| = 1

|\vec{b}| = 2

|\vec{c}| = 2

and each angle between pairs is

60^\circ

, find

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

Squared-length terms: $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 1 + 4 + 4 = 9$ .
Each cross term uses $\cos 60^\circ = \tfrac{1}{2}$ : $\vec{a}\cdot\vec{b} = 1\cdot 2\cdot\tfrac{1}{2} = 1$ , $\vec{b}\cdot\vec{c} = 2\cdot 2\cdot\tfrac{1}{2} = 2$ , $\vec{c}\cdot\vec{a} = 2\cdot 1\cdot\tfrac{1}{2} = 1$ .
So $2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 2(1 + 2 + 1) = 8$ .
$|\vec{a}+\vec{b}+\vec{c}|^2 = 9 + 8 = 17$ , so $|\vec{a}+\vec{b}+\vec{c}| = \sqrt{17}$ .

Answer:

|\vec{a} + \vec{b} + \vec{c}| = \sqrt{17}

Practice this conceptself-check · 4 quick reps

Try it yourself

Vectors

\vec{a}, \vec{b}, \vec{c}

have magnitudes 1, 2, 3. Given

\vec{a}\perp(\vec{b}+\vec{c})

\vec{b}\perp(\vec{c}+\vec{a})

\vec{c}\perp(\vec{a}+\vec{b})

, find

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

Practice — Level 1 (4 reps)

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

1.
If all three pairwise dot products are 0, $|\vec{a}+\vec{b}+\vec{c}|^2 = ?$
2.
$|\vec{a}|=1, |\vec{b}|=2, |\vec{c}|=2$ , all cross terms zero. $|\vec{a}+\vec{b}+\vec{c}|=?$
3.
What does $\vec{a}\perp\vec{b}$ make $\vec{a}\cdot\vec{b}$ equal?
4.
$\cos 60^\circ = ?$ (used in every angle version)

From the bank · past-year question

Example 4VectorsMODERATE

|\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5

and each angle between pairs is

60°

, then the value of

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

[Q104 · 14th May Shift 2 · 2024]

Add the cross terms with the factor of 2

The expansion is

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

. Forgetting the

2

halves every cross term — a frequent wrong answer when the angle version has non-zero dot products.

Perpendicularity conditions cancel ALL cross terms at once

The three conditions

\vec{a}\perp(\vec{b}+\vec{c})

etc. don't say each individual dot product is zero — they say their SUM is zero. That's all you need: the whole

2\sum(\vec{a}\cdot\vec{b})

term drops, leaving just

\sqrt{\sum|\cdot|^2}

Take the square root at the very end

You compute

|\vec{a}+\vec{b}+\vec{c}|^2

first. For magnitudes

1, 8, 4

with zero cross terms it's

81

, and the answer is

\sqrt{81} = 9

— NOT

81

. The distractor that leaves the squared value un-rooted is the classic trap.

Drill 3 more on magnitude of a sum from angles or perpendicularity

Concept 5 of 6

Magnitude of a sum with projection-equality and a perpendicular pair

Intuition

A second flavour of the same expansion. Instead of stating angles directly, the question gives an indirect handle: "the projection of

\vec{b}

along

\vec{a}

equals the projection of

\vec{c}

along

\vec{a}

" forces

(\vec{b}-\vec{c})\cdot\vec{a} = 0

, and "

\vec{b}\perp\vec{c}

" forces

\vec{b}\cdot\vec{c} = 0

. Feed both into the squared-magnitude expansion and the cross terms collapse.

Definition

Two facts unlock these:

**Equal projections along $\vec{a}$ :** projection of $\vec{b}$ on $\vec{a}$ is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}$ ; equating it to $\vec{c}$ 's projection gives $\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}$ , i.e. $(\vec{b}-\vec{c})\cdot\vec{a} = 0$ .
Perpendicular pair: $\vec{b}\perp\vec{c}\Rightarrow\vec{b}\cdot\vec{c} = 0$ .

Then expand the target, grouping the difference: $|\vec{a}+\vec{b}-\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}-\vec{c}|^2 + 2\vec{a}\cdot(\vec{b}-\vec{c})$ . The last term is 0 (equal projections), and $|\vec{b}-\vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2$ (perpendicular pair).

Grouped expansion with a perpendicular pair

|\vec{a} + \vec{b} - \vec{c}|^2 = |\vec{a}|^2 + |\vec{b} - \vec{c}|^2 + 2\,\vec{a}\cdot(\vec{b} - \vec{c})

$\vec{a}\cdot(\vec{b}-\vec{c}) = 0$ from equal projections of $\vec{b}, \vec{c}$ along $\vec{a}$
$|\vec{b}-\vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2$ from $\vec{b}\perp\vec{c}$

Worked example

Let

|\vec{a}| = 3

|\vec{b}| = 6

|\vec{c}| = 6

. If the projection of

\vec{b}

\vec{a}

equals the projection of

\vec{c}

\vec{a}

, and

\vec{b}\perp\vec{c}

, find

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

Equal projections: $\vec{a}\cdot(\vec{b} - \vec{c}) = 0$ , so the cross term with $\vec{a}$ drops.
$\vec{b}\perp\vec{c}$ : $|\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 = 36 + 36 = 72$ .
Combine: $|\vec{a}+\vec{b}-\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}-\vec{c}|^2 = 9 + 72 = 81$ .
$|\vec{a}+\vec{b}-\vec{c}| = \sqrt{81} = 9$ .

Answer:

|\vec{a} + \vec{b} - \vec{c}| = 9

Practice this conceptself-check · 4 quick reps

Try it yourself

Vectors

\vec{u}, \vec{v}, \vec{w}

have magnitudes 1, 2, 3. The projection of

\vec{v}

along

\vec{u}

equals that of

\vec{w}

along

\vec{u}

, and

\vec{v}\perp\vec{w}

. Find

|\vec{u} - \vec{v} + \vec{w}|

Practice — Level 1 (4 reps)

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

1.
Equal projections of $\vec{b}, \vec{c}$ along $\vec{a}$ give which dot-product equation?
2.
If $\vec{b}\perp\vec{c}$ , then $|\vec{b}-\vec{c}|^2 = ?$
3.
$|\vec{a}|=2, |\vec{b}|=4, |\vec{c}|=4$ , all relevant cross terms zero. $|\vec{a}+\vec{b}-\vec{c}|^2 = ?$
4.
$|\vec{a}+\vec{b}-\vec{c}|^2 = 36$ gives $|\vec{a}+\vec{b}-\vec{c}| = ?$

From the bank · past-year question

Example 5VectorsHARD

Let the vectors

\vec{a},\vec{b},\vec{c}

be such that

|\vec{a}|=2, |\vec{b}|=4

and

|\vec{c}|=4

. If the projection of

\vec{b}

\vec{a}

is equal to the projection of

\vec{c}

\vec{a}

and

\vec{b}

is perpendicular to

\vec{c}

, then the value of

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

[Q107 · 16th May Shift 1 · 2023]

"Equal projections along $\vec{a}$ " means $(\vec{b}-\vec{c})\cdot\vec{a} = 0$ , not $\vec{b} = \vec{c}$

The projections being equal only forces

\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}

—

\vec{b}

and

\vec{c}

can still differ wildly. Use it to kill exactly the

\vec{a}

-cross term, nothing more.

Watch the sign on the vector you subtract

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

, the

\vec{c}

term is subtracted, but

|\vec{b}-\vec{c}|^2

still ADDS the squared lengths when

\vec{b}\perp\vec{c}

(the cross term

-2\vec{b}\cdot\vec{c}

is zero). The minus sign only matters through the dot product, which vanishes here.

Drill 3 more on magnitude of a sum with projection-equality and a perpendicular pair

Concept 6 of 6

Unit vector parallel to a parallelogram's diagonal

Intuition

If two adjacent sides of a parallelogram are

\vec{a}

and

\vec{b}

, one diagonal runs from the shared vertex to the opposite corner — and by the triangle/parallelogram law that diagonal IS

\vec{a} + \vec{b}

. To get a unit vector along it, add the sides, then divide by the resulting magnitude.

Definition

For a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ from a common vertex:

The diagonal through that vertex is $\vec{d} = \vec{a} + \vec{b}$ .
The other diagonal is $\vec{a} - \vec{b}$ (or $\vec{b} - \vec{a}$ ).

The unit vector parallel to the diagonal is $\hat{d} = \dfrac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}$ .

Unit vector along a diagonal

\hat{d} = \dfrac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}

$\vec{a}, \vec{b}$ adjacent sides from the shared vertex
$\vec{a} + \vec{b}$ the diagonal through that vertex

Visualization · add two vectors tip-to-tail

a₁: 4a₂: 1b₁: 1b₂: 3

a + b = (5, 4)|a + b| = 6.40|a| + |b| = 7.29

Worked example

Two adjacent sides of a parallelogram are

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

and

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

. Find the unit vector parallel to the diagonal

\vec{a} + \vec{b}

Add the sides: $\vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} - 4\hat{k}$ .
Magnitude: $|\vec{a} + \vec{b}| = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3}$ .
Unit vector: $\dfrac{4\hat{i} + 4\hat{j} - 4\hat{k}}{4\sqrt{3}} = \dfrac{1}{\sqrt{3}}(\hat{i} + \hat{j} - \hat{k})$ .

Answer:

\dfrac{1}{\sqrt{3}}(\hat{i} + \hat{j} - \hat{k})

Practice this conceptself-check · 4 quick reps

Try it yourself

Adjacent sides of a parallelogram are

\hat{i} + \hat{j}

and

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

. Find the unit vector along the diagonal

\vec{a} + \vec{b}

Practice — Level 1 (4 reps)

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

1.
Sides $\vec{a}, \vec{b}$ : which diagonal is $\vec{a} + \vec{b}$ ?
2.
Sides $2\hat{i}$ and $2\hat{j}$ : diagonal $\vec{a}+\vec{b}$ and its magnitude?
3.
The OTHER diagonal of a parallelogram with sides $\vec{a}, \vec{b}$ ?
4.
Unit vector along $6\hat{i} + 8\hat{j}$ ?

From the bank · past-year question

Example 6VectorsMODERATE

Two adjacent sides of a parallelogram are

2\hat{i}-4\hat{j}+5\hat{k}

and

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

, then the unit vector parallel to its diagonal is

[Q134 · 9th May Shift 2 · 2024]

Diagonal $\vec{a}+\vec{b}$ , not $\vec{a}-\vec{b}$

A parallelogram has two diagonals:

\vec{a} + \vec{b}

(through the shared vertex) and

\vec{a} - \vec{b}

(the other one). The question usually wants the SUM; reading it as the difference gives a different unit vector entirely.

Normalise the diagonal's own magnitude, not a stray $\sqrt{77}$

Compute

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

AFTER adding — for

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

that is

\sqrt{9+36+4} = 7

. Distractors often divide by an unrelated magnitude (like

\sqrt{77}

from

\sqrt{49 + 28}

); always re-square the summed components.

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 (6)

Vectors and component form
Component form
$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$
Magnitude of a vector and distance between two points
Magnitude and distance
$|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \qquad AB = |\vec{b} - \vec{a}|$
Unit vector along a given direction
Unit vector
$\hat{a} = \dfrac{\vec{a}}{|\vec{a}|} \qquad r\,\hat{a} = \dfrac{r}{|\vec{a}|}\,\vec{a}$
Magnitude of a sum from angles or perpendicularity
Magnitude of a sum (squared)
$|\vec{a} + \vec{b} + \vec{c}|^2 = \sum|\vec{a}|^2 + 2\sum(\vec{a}\cdot\vec{b}), \qquad \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$
Magnitude of a sum with projection-equality and a perpendicular pair
Grouped expansion with a perpendicular pair
$|\vec{a} + \vec{b} - \vec{c}|^2 = |\vec{a}|^2 + |\vec{b} - \vec{c}|^2 + 2\,\vec{a}\cdot(\vec{b} - \vec{c})$
Unit vector parallel to a parallelogram's diagonal
Unit vector along a diagonal
$\hat{d} = \dfrac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}$

Watch out for (11)

A missing axis means a zero component, not a 2-D vector
→ Vectors and component form
$\overrightarrow{AB} = \vec{b} - \vec{a}$ — head minus tail
→ Magnitude of a vector and distance between two points
Magnitude needs every component squared
→ Magnitude of a vector and distance between two points
Divide by the magnitude, don't subtract it
→ Unit vector along a given direction
Add the cross terms with the factor of 2
→ Magnitude of a sum from angles or perpendicularity
Perpendicularity conditions cancel ALL cross terms at once
→ Magnitude of a sum from angles or perpendicularity
Take the square root at the very end
→ Magnitude of a sum from angles or perpendicularity
"Equal projections along $\vec{a}$ " means $(\vec{b}-\vec{c})\cdot\vec{a} = 0$ , not $\vec{b} = \vec{c}$
→ Magnitude of a sum with projection-equality and a perpendicular pair
Watch the sign on the vector you subtract
→ Magnitude of a sum with projection-equality and a perpendicular pair
Diagonal $\vec{a}+\vec{b}$ , not $\vec{a}-\vec{b}$
→ Unit vector parallel to a parallelogram's diagonal
Normalise the diagonal's own magnitude, not a stray $\sqrt{77}$
→ Unit vector parallel to a parallelogram's diagonal

Mastery check — 5 interleaved questions

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

Example 1VectorsMODERATE

Let

\vec{a}, \vec{b}

and

\vec{c}

be vectors of magnitude 2, 3 and 4 respectively. If

\vec{a}

is perpendicular to

(\vec{b}+\vec{c})

\vec{b}

is perpendicular to

(\vec{c}+\vec{a})

and

\vec{c}

is perpendicular to

(\vec{a}+\vec{b})

, then the magnitude of

\vec{a}+\vec{b}+\vec{c}

is equal to

[Q108 · 9th May Shift 2 · 2023]

Example 2VectorsHARD

Let the vectors

\vec{a},\vec{b},\vec{c}

be such that

|\vec{a}|=2,|\vec{b}|=4,|\vec{c}|=4

. If the projection of

\vec{b}

\vec{a}

equals the projection of

\vec{c}

\vec{a}

and

\vec{b}

is perpendicular to

\vec{c}

, then the value of

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

is equal to

[Q139 · 3rd May 2nd Shift · 2023]

Example 3VectorsMODERATE

Let

\vec{A},\vec{B},\vec{C}

be vectors of lengths 3 units, 4 units, 5 units respectively. Let

\vec{A}\perp(\vec{B}+\vec{C})

\vec{B}\perp(\vec{C}+\vec{A})

and

\vec{C}\perp(\vec{A}+\vec{B})

, then the length of vector

\vec{A}+\vec{B}+\vec{C}

[Q139 · 10th May Shift 1 · 2023]

Example 4VectorsHARD

Let

\vec{u}, \vec{v}, \vec{w}

be vectors with

|\vec{u}|=1, |\vec{v}|=2, |\vec{w}|=3

. Projection of

\vec{v}

along

\vec{u}

equals projection of

\vec{w}

along

\vec{u}

, and

\vec{v} \perp \vec{w}

. Then

|\vec{u} - \vec{v} + \vec{w}|

equals

[Q110 · 14th May Shift 2 · 2024]

Example 5VectorsMODERATE

\vec{a}, \vec{b}, \vec{c}

are three vectors such that

\vec{a}\cdot(\vec{b}+\vec{c}) + \vec{b}\cdot(\vec{c}+\vec{a}) + \vec{c}\cdot(\vec{a}+\vec{b}) = 0

and

|\vec{a}|=1

|\vec{b}|=8

and

|\vec{c}|=4

, then

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

has the value

[Q124 · 11th May Shift 1 · 2024]

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Related notes

Vector algebra fundamentals →