MHT-CET Maths · Vectors

Dot Product, Angle, and Perpendicularity

The scalar a·b = |a||b|cosθ that measures alignment — the engine behind angle, perpendicularity, projection, and direction-cosine questions.

All Vectors notes MHT-CET Maths strategy

Why this matters

The dot product collapses two vectors into one number that encodes their angle, so almost every Vectors question in MHT-CET routes through it: find the angle, test perpendicularity, solve for a parameter that makes two vectors perpendicular, or project one vector onto another. This is the single biggest Vectors subtopic — 35 PYQs across 2017–2026 with roughly 23% HARD — and the staples are the perpendicular-parameter setup (find λ so a+λb ⊥ c), the unit-vector constraint angle, and projection of a segment onto a line. Master the perpendicularity test (a·b = 0), the angle formula, and the expand-the-constraint workflow and you have most of the chapter's marks.

Concept 1 of 15

Dot product — the two faces of a·b

Intuition

The dot product turns two vectors into a single number. It has two faces that you switch between constantly: a GEOMETRIC face

|\vec a||\vec b|\cos\theta

(use it when angles or magnitudes are given) and an ALGEBRAIC face

a_1b_1+a_2b_2+a_3b_3

(use it when components are given). Both give the same scalar.

Definition

For $\vec a = a_1\hat i + a_2\hat j + a_3\hat k$ and $\vec b = b_1\hat i + b_2\hat j + b_3\hat k$ :

Geometric: $\vec a\cdot\vec b = |\vec a|\,|\vec b|\cos\theta$ , where $\theta$ is the angle between them ( $0\le\theta\le\pi$ ).
Component: $\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$ .
It is commutative ( $\vec a\cdot\vec b = \vec b\cdot\vec a$ ), distributive over addition, and always a scalar.
Self dot: $\vec a\cdot\vec a = |\vec a|^2$ . Basis: $\hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1$ and $\hat i\cdot\hat j = \hat j\cdot\hat k = \hat k\cdot\hat i = 0$ .

Dot product — geometric and component forms

\vec a\cdot\vec b = |\vec a|\,|\vec b|\cos\theta = a_1 b_1 + a_2 b_2 + a_3 b_3

$\theta$ angle between the vectors, in $[0,\pi]$
$a_i, b_i$ components along $\hat i, \hat j, \hat k$
$\vec a\cdot\vec a$ equals $|\vec a|^2$

Diagram · work = F · d = |F||d| cos θ

Only the part of the force along the displacement does work: W = F · d = |F||d| cos θ. A force perpendicular to the motion (θ = 90°) does zero work; one opposing it (θ > 90°) does negative work.

Worked example

Evaluate

\vec a\cdot\vec b

for

\vec a = 3\hat i - \hat j + 2\hat k

and

\vec b = \hat i + 4\hat j - \hat k

Use the component form: multiply matching components and add.
$\vec a\cdot\vec b = (3)(1) + (-1)(4) + (2)(-1) = 3 - 4 - 2$ .
Sum: $\vec a\cdot\vec b = -3$ .

Answer:

\vec a\cdot\vec b = -3

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$\vec a\cdot\vec b$ for $\vec a = \hat i + 2\hat j$ , $\vec b = 3\hat i + \hat j$ ?
2.
$\hat i\cdot\hat j = ?$
3.
$\hat k\cdot\hat k = ?$
4.
For $\vec a = 2\hat i - \hat j + \hat k$ , $\vec a\cdot\vec a = ?$

Dot product is a scalar — never a vector

An MCQ option that returns

\hat i, \hat j, \hat k

components for

\vec a\cdot\vec b

is wrong on type grounds alone. The dot product is always a single number; the cross product is the one that gives a vector.

Match components in the SAME direction only

\vec a\cdot\vec b

multiplies

a_1b_1 + a_2b_2 + a_3b_3

— the

\hat i

of one with the

\hat i

of the other. Multiplying

a_1 b_2

(an

\hat i

with a

\hat j

) is the most common slip, because those mixed products are exactly the zero terms.

Concept 2 of 15

Magnitude of a combination via the dot product

Intuition

The magnitude of any vector combination is found by squaring it as a dot product with itself. When the vectors are in component form, build the combination component-by-component first, then take the square-root of the sum of squares. When only magnitudes and angles are given, expand

|\vec a\pm\vec b|^2

into self-dots plus a cross term.

Definition

For a combination written in components, $|\vec v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ . Abstractly, $|\,p\vec a + q\vec b\,|^2 = p^2|\vec a|^2 + 2pq\,(\vec a\cdot\vec b) + q^2|\vec b|^2$ . This is just $(p\vec a + q\vec b)\cdot(p\vec a + q\vec b)$ expanded.

Magnitude of a linear combination

|p\vec a + q\vec b|^2 = p^2|\vec a|^2 + 2pq\,(\vec a\cdot\vec b) + q^2|\vec b|^2

$p, q$ scalar coefficients
$\vec a\cdot\vec b$ the only cross term — vanishes if $\vec a\perp\vec b$

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

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

and

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

, find

|2\vec a + \vec b|

Build the combination component-by-component: $2\vec a = 2\hat i + 4\hat j - 2\hat k$ .
$2\vec a + \vec b = (2+3)\hat i + (4-1)\hat j + (-2+2)\hat k = 5\hat i + 3\hat j + 0\hat k$ .
Magnitude: $|2\vec a + \vec b| = \sqrt{5^2 + 3^2 + 0^2} = \sqrt{34}$ .

Answer:

|2\vec a + \vec b| = \sqrt{34}

Practice this conceptself-check · 4 quick reps

Try it yourself

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

and

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

, find

|\vec a - \vec b|

Practice — Level 1 (4 reps)

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

1.
$|3\hat i + 4\hat k|$ ?
2.
$\vec a = \hat i + \hat j$ , $\vec b = \hat i - \hat j$ : $|\vec a + \vec b|$ ?
3.
Unit perpendicular $\vec a, \vec b$ : $|\vec a - \vec b|$ ?
4.
$|\hat i - 2\hat j + 2\hat k|$ ?

From the bank · past-year question

Example 2VectorsEASY

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

and

\vec{b} = 2\hat{i} - 4\hat{j} - 3\hat{k}

, then

|\vec{a} - 2\vec{b}|

will be

[Q133 · May Shift 1 · 2021]

Scale FIRST, then subtract — watch the double minus

|\vec a - 2\vec b|

, the

\hat j

-component becomes

a_2 - 2b_2

. If

b_2

is itself negative,

-2b_2

is positive — e.g.

a_2 = -2, b_2 = -4

gives

-2 - 2(-4) = 6

, not

-10

. The sign trap lives in the doubled, negative component.

Magnitude is never negative

|\vec v| = \sqrt{\,\cdot\,}

takes the positive root. If a derivation yields a negative number under the root or a negative final magnitude, the arithmetic is wrong — recompute the components.

Drill 1 more on magnitude of a combination via the dot product

Concept 3 of 15

Perpendicularity test and solving for a parameter

Intuition

Two non-zero vectors are perpendicular exactly when their dot product is zero. The bank's single most-repeated question hands you

\vec a + \lambda\vec b

(a vector with an unknown

\lambda

) and a fixed

\vec c

, says the two are perpendicular, and asks for

\lambda

. The recipe never changes: write

(\vec a+\lambda\vec b)\cdot\vec c = 0

, expand, and solve one linear equation.

Definition

For non-zero $\vec a, \vec b$ : $\vec a\perp\vec b \iff \vec a\cdot\vec b = 0$ . To find $\lambda$ such that $(\vec a + \lambda\vec b)\perp\vec c$ : set $(\vec a + \lambda\vec b)\cdot\vec c = 0$ , which gives $\vec a\cdot\vec c + \lambda(\vec b\cdot\vec c) = 0$ , hence $\lambda = -\dfrac{\vec a\cdot\vec c}{\vec b\cdot\vec c}$ . Remember a missing third component (e.g. $\vec c = 3\hat i + \hat j$ ) means its $\hat k$ -component is 0, not ignored.

Perpendicularity and the perp-parameter

\vec a\perp\vec b \iff \vec a\cdot\vec b = 0 \qquad (\vec a + \lambda\vec b)\perp\vec c \Rightarrow \lambda = -\frac{\vec a\cdot\vec c}{\vec b\cdot\vec c}

$\lambda$ the unknown scalar to solve for
$\vec a\cdot\vec c,\ \vec b\cdot\vec c$ two scalar dot products, computed once each

Worked example

Find

\lambda

so that

\vec a + \lambda\vec b

is perpendicular to

\vec c

, where

\vec a = \hat i + \hat j + 4\hat k

\vec b = 2\hat i - \hat j + \hat k

\vec c = 2\hat i + \hat j

Compute $\vec a\cdot\vec c = (1)(2) + (1)(1) + (4)(0) = 3$ (note $\vec c$ has $\hat k$ -component 0).
Compute $\vec b\cdot\vec c = (2)(2) + (-1)(1) + (1)(0) = 3$ .
Set $\vec a\cdot\vec c + \lambda(\vec b\cdot\vec c) = 0$ : $3 + 3\lambda = 0$ .
Solve: $\lambda = -1$ .

Answer:

\lambda = -1

Practice this conceptself-check · 4 quick reps

Try it yourself

Find

\lambda

so that

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

and

\hat i + \lambda\hat j - 5\hat k

are perpendicular.

Practice — Level 1 (4 reps)

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

1.
Are $\hat i + \hat j$ and $\hat i - \hat j$ perpendicular?
2.
$\lambda$ for $\hat i + \lambda\hat j$ perpendicular to $2\hat i - \hat j$ ?
3.
Perpendicularity condition for $\vec a, \vec b$ ?
4.
$\hat k$ -component of $\vec c = 3\hat i + \hat j$ ?

From the bank · past-year question

Example 3VectorsEASY

\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 vectors such that

\vec{a}+\lambda\vec{b}

is perpendicular to

\vec{c}

, then value of

\lambda

[Q131 · 16th May Shift 2 · 2023]

A missing component is ZERO, not absent

When

\vec c = 3\hat i + \hat j

, its

\hat k

-component is

0

. In the dot product that term contributes

0

— but you must still write the slot so the

\hat i

and

\hat j

terms line up with the right components of the other vector.

Solve for $\lambda$ cleanly: $\lambda = -(\vec a\cdot\vec c)/(\vec b\cdot\vec c)$

After expanding, the equation is always linear in

\lambda

(\vec a\cdot\vec c) + \lambda(\vec b\cdot\vec c) = 0

. Compute the two dot products as numbers first, then divide. Sign errors creep in when people expand all the components at once instead.

Order matters: $\vec a + \lambda\vec b$ vs $\vec b + \lambda\vec a$

If the question reads

\vec b + \lambda\vec a

perpendicular to

\vec c

, then

\lambda = -(\vec b\cdot\vec c)/(\vec a\cdot\vec c)

— the roles flip. Read which vector carries the

\lambda

before substituting.

Drill 5 more on perpendicularity test and solving for a parameter

Concept 4 of 15

Disguised perpendicularity — equal-diagonal and Pythagoras forms

Intuition

Perpendicularity hides behind two algebraic disguises that exams love.

|\vec a + \vec b| = |\vec a - \vec b|

says the two diagonals of the parallelogram on

\vec a, \vec b

are equal — which happens exactly when it is a rectangle, i.e.

\vec a \perp \vec b

. And

|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2

is just the Pythagoras identity, again forcing

\vec a\cdot\vec b = 0

Definition

For non-zero $\vec a, \vec b$ , the following are all equivalent to $\vec a \perp \vec b$ :

$\vec a\cdot\vec b = 0$
$|\vec a + \vec b| = |\vec a - \vec b|$ (equal diagonals)
$|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2$ (Pythagoras)

Equivalent perpendicularity statements

\vec a\perp\vec b \iff |\vec a + \vec b| = |\vec a - \vec b| \iff |\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2

$|\vec a \pm \vec b|$ lengths of the two parallelogram diagonals

Worked example

|\vec a + \vec b| = |\vec a - \vec b|

, find the angle between

\vec a

and

\vec b

Square both sides: $|\vec a|^2 + 2\vec a\cdot\vec b + |\vec b|^2 = |\vec a|^2 - 2\vec a\cdot\vec b + |\vec b|^2$ .
Cancel the like terms: $4\,\vec a\cdot\vec b = 0$ , so $\vec a\cdot\vec b = 0$ .
A zero dot product (non-zero vectors) means the vectors are perpendicular.

Answer:

\theta = \dfrac{\pi}{2}

(i.e.

90^\circ

)

Practice this conceptself-check · 4 quick reps

Try it yourself

Vectors

\vec a, \vec b

satisfy

|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2

. What is

\vec a\cdot\vec b

Practice — Level 1 (4 reps)

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

1.
$|\vec a + \vec b| = |\vec a - \vec b|$ implies which angle?
2.
Expand $|\vec a - \vec b|^2$ .
3.
Equal parallelogram diagonals mean the shape is a ___?
4.
If $|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2$ , then $\vec a\cdot\vec b = ?$

$|\vec a + \vec b| = |\vec a - \vec b|$ is NOT $\vec a = \vec b$

It is tempting to read equal magnitudes as equal vectors. Squaring shows it collapses to

\vec a\cdot\vec b = 0

— a perpendicularity condition, not an equality. Geometrically, the diagonals of a parallelogram are equal iff it is a rectangle.

Pythagoras only works on the PLUS combination for a right angle

|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2

signals perpendicularity. Don't confuse it with

|\vec a + \vec b|^2 = |\vec a|^2 + 2\vec a\cdot\vec b + |\vec b|^2

, which is the general expansion that holds for any angle.

Concept 5 of 15

Angle between two vectors via cosθ

Intuition

Rearrange

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

and the angle drops out: divide the dot product by the product of magnitudes. The sign of the result tells you the type of angle immediately — positive means acute, negative means obtuse, and zero means perpendicular.

Definition

For non-zero $\vec a, \vec b$ at angle $\theta \in [0,\pi]$ : $\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|}$ . Sign rule: $\vec a\cdot\vec b > 0 \Rightarrow$ acute; $\vec a\cdot\vec b = 0 \Rightarrow$ right angle; $\vec a\cdot\vec b < 0 \Rightarrow$ obtuse. The same formula works on combinations: to find the angle between $3\vec a + 5\vec b$ and $5\vec a + 3\vec b$ , build each combination in components first, then apply the formula.

Angle from the dot product

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

$\theta$ angle between the vectors, in $[0,\pi]$
$|\vec a|, |\vec b|$ magnitudes (always positive)

Visualization · project a onto b

a₁: 3a₂: 4b₁: 6b₂: 0

a·b = 18proj = (a·b)/|b| = 3.00θ ≈ 53°

The green band is how far a reaches along b — its signed length is the scalar projection (a·b)/|b|. Push the angle past 90° and the dot product, and the projection, turn negative.

Worked example

Find the angle between

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

and

\vec b = \hat i - \hat j

Dot product: $\vec a\cdot\vec b = (1)(1) + (1)(-1) + (1)(0) = 0$ .
A zero dot product with non-zero vectors means they are perpendicular.
Hence $\theta = \dfrac{\pi}{2}$ — no need to compute magnitudes.

Answer:

\theta = \dfrac{\pi}{2}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the angle between

\vec a = \hat i + \hat j

and

\vec b = \hat i

Practice — Level 1 (4 reps)

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

1.
$\cos\theta$ when $\vec a\cdot\vec b = 6$ , $|\vec a| = 3$ , $|\vec b| = 4$ ?
2.
If $\cos\theta < 0$ , the angle is?
3.
Angle between $\vec a$ and $\vec a$ ?
4.
Angle between $\hat i$ and $\hat k$ ?

From the bank · past-year question

Example 5VectorsMODERATE

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

and

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

are two vectors, then the angle between the vectors

3\vec{a} + 5\vec{b}

and

5\vec{a} + 3\vec{b}

[Q114 · Shift 1 · 2023]

Build the combinations BEFORE taking the angle

For the angle between

3\vec a + 5\vec b

and

5\vec a + 3\vec b

, you must first compute each combination's components, then dot and divide. You cannot shortcut by 'mixing' the angle between

\vec a

and

\vec b

directly.

Leave the answer as $\cos^{-1}(\cdot)$ when it is not a standard angle

\cos\theta = \tfrac{13}{19}

, the angle is

\cos^{-1}\tfrac{13}{19}

— not a round number. The correct option is the inverse-cosine form; don't force it into

\pi/3

\pi/4

Drill 1 more on angle between two vectors via cosθ

Concept 6 of 15

Angle from a unit-vector perpendicularity constraint

Intuition

A whole family of MODERATE PYQs gives two unit vectors and says some combination — like

\vec a + 2\vec b

and

5\vec a - 4\vec b

— is perpendicular, then asks for the angle between

\vec a

and

\vec b

. The workflow never changes: expand the perpendicularity dot product, replace

|\vec a|^2 = |\vec b|^2 = 1

(unit vectors) and

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

, then solve one linear equation for

\cos\theta

Definition

Given $(\alpha\vec a + \beta\vec b)\cdot(\gamma\vec a + \delta\vec b) = 0$ with $\vec a, \vec b$ unit vectors, expand to $\alpha\gamma|\vec a|^2 + (\alpha\delta + \beta\gamma)\,\vec a\cdot\vec b + \beta\delta|\vec b|^2 = 0$ . Substitute $|\vec a|^2 = |\vec b|^2 = 1$ and $\vec a\cdot\vec b = \cos\theta$ : $\alpha\gamma + (\alpha\delta + \beta\gamma)\cos\theta + \beta\delta = 0$ , then read off $\cos\theta$ .

Expansion of a perpendicular constraint (unit vectors)

(\alpha\vec a + \beta\vec b)\cdot(\gamma\vec a + \delta\vec b) = \alpha\gamma + (\alpha\delta + \beta\gamma)\cos\theta + \beta\delta = 0

$\alpha,\beta,\gamma,\delta$ given coefficients in the two combinations
$\cos\theta$ equals $\vec a\cdot\vec b$ — the unknown to isolate

Worked example

\vec a, \vec b

are unit vectors such that

3\vec a + \vec b

and

\vec a - 3\vec b

are perpendicular. Find the angle between

\vec a

and

\vec b

Set $(3\vec a + \vec b)\cdot(\vec a - 3\vec b) = 0$ .
Expand: $3|\vec a|^2 + (-9 + 1)\vec a\cdot\vec b - 3|\vec b|^2 = 0$ .
Unit vectors: $3(1) - 8\cos\theta - 3(1) = 0$ , so $-8\cos\theta = 0$ .
$\cos\theta = 0 \Rightarrow \theta = \dfrac{\pi}{2}$ .

Answer:

\theta = \dfrac{\pi}{2}

Practice this conceptself-check · 4 quick reps

Try it yourself

\vec a, \vec b

are unit vectors and

\vec a + \vec b

is perpendicular to

\vec a - \vec b

. Find the angle between

\vec a

and

\vec b

Practice — Level 1 (4 reps)

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

1.
Unit $\vec a, \vec b$ : expand $(\vec a + 2\vec b)\cdot(\vec a - 2\vec b)$ .
2.
If $5 + 6\cos\theta - 8 = 0$ , find $\cos\theta$ .
3.
$\cos\theta = \tfrac{1}{2}$ gives which angle?
4.
$\cos\theta = -\tfrac{1}{2}$ gives which angle?

From the bank · past-year question

Example 6VectorsMODERATE

\vec{a}

and

\vec{b}

are two unit vectors such that

\vec{a}+2\vec{b}

and

5\vec{a}-4\vec{b}

are perpendicular to each other, then the angle between

\vec{a}

and

\vec{b}

[Q149 · 10th May Shift 2 · 2024]

Coefficient swap flips the angle: check WHICH combination

(\vec a + 2\vec b)\cdot(5\vec a - 4\vec b) = 0

gives

\cos\theta = +\tfrac{1}{2}

(so

\tfrac{\pi}{3}

), but

(5\vec a + 4\vec b)\cdot(\vec a - 2\vec b) = 0

gives

\cos\theta = -\tfrac{1}{2}

(so

\tfrac{2\pi}{3}

). The sign of the cross-term coefficient decides acute vs obtuse — read the coefficients exactly.

Unit vectors mean $|\vec a|^2 = 1$ , not $0$

The self-dot terms

|\vec a|^2

and

|\vec b|^2

become

1

, not

0

. Dropping them (as if

\vec a

were the zero vector) loses the constant terms and gives a wrong

\cos\theta

Don't drop a cross term — there are TWO middle products

(\alpha\vec a + \beta\vec b)\cdot(\gamma\vec a + \delta\vec b)

has FOUR products; the two middle ones (

\alpha\delta

and

\beta\gamma

) both contribute to the

\vec a\cdot\vec b

coefficient. A factor-of-2 wrong answer usually means one was dropped.

Drill 5 more on angle from a unit-vector perpendicularity constraint

Concept 7 of 15

Scalar and vector projection

Intuition

The scalar projection of

\vec a

\vec b

is the signed length of

\vec a

's shadow along

\vec b

: divide the dot product by

|\vec b|

. The vector projection re-attaches the direction by multiplying that scalar by the unit vector

\hat b

. A huge slice of CET projection PYQs are 'projection of

\overrightarrow{AB}

\overrightarrow{CD}

' — build both difference vectors from the points first.

Definition

Scalar projection of $\vec a$ on $\vec b$ : $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ (a signed number). Vector projection of $\vec a$ on $\vec b$ : $\left(\dfrac{\vec a\cdot\vec b}{|\vec b|^2}\right)\vec b$ (a vector along $\vec b$ ). For a segment on a line, set $\vec a = \overrightarrow{AB} = B - A$ and $\vec b = \overrightarrow{CD} = D - C$ (or the line's direction ratios), then apply the formula.

Scalar and vector projection of a on b

\text{scalar} = \frac{\vec a\cdot\vec b}{|\vec b|}, \qquad \text{vector} = \frac{\vec a\cdot\vec b}{|\vec b|^2}\,\vec b

$\vec a\cdot\vec b$ dot product (carries the sign)
$|\vec b|$ divide once for scalar; squared for vector

Visualization · project a onto b

a₁: 3a₂: 4b₁: 6b₂: 0

a·b = 18proj = (a·b)/|b| = 3.00θ ≈ 53°

The green band is how far a reaches along b — its signed length is the scalar projection (a·b)/|b|. Push the angle past 90° and the dot product, and the projection, turn negative.

Worked example

Find the scalar projection of

\vec a = 2\hat i + 3\hat j + 2\hat k

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

Dot product: $\vec a\cdot\vec b = (2)(1) + (3)(2) + (2)(2) = 2 + 6 + 4 = 12$ .
Magnitude of $\vec b$ : $|\vec b| = \sqrt{1 + 4 + 4} = 3$ .
Scalar projection $= \dfrac{12}{3} = 4$ .

Answer:Scalar projection

= 4

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the projection of

\overrightarrow{AB}

\overrightarrow{CD}

where

A(1,0,0)

B(4,4,0)

C(0,0,0)

D(0,0,5)

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.
Projection of $\vec a\cdot\vec b = 6$ onto $\vec b$ with $|\vec b| = 2$ ?
3.
$\overrightarrow{AB}$ for $A(2,3,-1)$ , $B(-2,-4,3)$ ?
4.
$|\overrightarrow{CD}|$ for direction $(3,-6,2)$ ?

From the bank · past-year question

Example 7VectorsMODERATE

What will be the projection of the vector

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

on the line joining the points

(2,3,-1)

and

(-2,-4,3)

[Q132 · May Shift 1 · 2021]

Divide by $|\vec b|$ — projection is NOT just the dot product

The most common projection error is reporting

\vec a\cdot\vec b

and forgetting to divide by the magnitude of the vector you project ONTO. Scalar projection

= (\vec a\cdot\vec b)/|\vec b|

; the magnitude in the denominator is non-negotiable.

Scalar projection can be negative; magnitude of projection is its absolute value

If a question asks for the 'magnitude of the projection', take

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

. A negative scalar projection (e.g.

-\tfrac{1}{7}

) just means the shadow points opposite to

\vec b

— its magnitude is

\tfrac{1}{7}

Vector projection divides by $|\vec b|^2$ , then multiplies by $\vec b$

Don't confuse the two: scalar uses

|\vec b|^1

and gives a number; vector uses

|\vec b|^2

and multiplies by

\vec b

to give a vector. Writing

\tfrac{\vec a\cdot\vec b}{|\vec b|}\,\vec b

is dimensionally wrong.

Drill 4 more on scalar and vector projection

Concept 8 of 15

Projection onto the normal of a plane

Intuition

A trickier projection question asks for the projection of

\vec v

onto 'the vector perpendicular to the plane containing

\vec p

and

\vec q

'. That perpendicular direction is the cross product

\vec n = \vec p \times \vec q

. Once you have

\vec n

, it is an ordinary scalar projection of

\vec v

onto

\vec n

Definition

The vector perpendicular to the plane of $\vec p$ and $\vec q$ is the normal $\vec n = \vec p \times \vec q$ . The magnitude of the projection of $\vec v$ on this normal is $\dfrac{|\vec v\cdot\vec n|}{|\vec n|}$ . So the recipe is: cross-product to get $\vec n$ , then project $\vec v$ onto $\vec n$ by the usual scalar-projection rule.

Projection on the plane normal

\vec n = \vec p \times \vec q, \qquad \text{proj} = \frac{|\vec v\cdot\vec n|}{|\vec n|}

$\vec n = \vec p\times\vec q$ normal to the plane of $\vec p, \vec q$
$\vec v\cdot\vec n$ dot of the target vector with the normal

Worked example

Find the magnitude of the projection of

\vec v = \hat i + \hat j + \hat k

on the normal to the plane containing

\vec p = \hat i + \hat j

and

\vec q = \hat j + \hat k

Normal: $\vec n = \vec p \times \vec q = \begin{vmatrix}\hat i & \hat j & \hat k\\1 & 1 & 0\\0 & 1 & 1\end{vmatrix} = \hat i(1) - \hat j(1) + \hat k(1) = \hat i - \hat j + \hat k$ .
$|\vec n| = \sqrt{1 + 1 + 1} = \sqrt{3}$ .
$\vec v\cdot\vec n = (1)(1) + (1)(-1) + (1)(1) = 1$ .
Projection magnitude $= \dfrac{|1|}{\sqrt{3}} = \dfrac{1}{\sqrt{3}}$ .

Answer:Projection

= \dfrac{1}{\sqrt{3}}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
Vector perpendicular to the plane of $\vec p, \vec q$ ?
2.
$\vec n = (\hat i + \hat j + \hat k)\times(\hat i + 2\hat j + 3\hat k) = ?$
3.
$|\hat i - 2\hat j + \hat k|$ ?
4.
Projection magnitude of $\vec v$ on $\vec n$ formula?

From the bank · past-year question

Example 8VectorsMODERATE

The magnitude of the projection of the vector

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

on the vector perpendicular to the plane containing the vectors

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

and

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

[Q120 · 9th May Shift 2 · 2024]

'Perpendicular to the plane' means CROSS product, not dot

The phrase 'vector perpendicular to the plane containing

\vec p

and

\vec q

' is asking for

\vec p \times \vec q

. Trying to project onto

\vec p

\vec q

directly answers a different question.

Project onto the NORMAL, then divide by $|\vec n|$

After computing

\vec n

, it is still an ordinary scalar projection:

(\vec v\cdot\vec n)/|\vec n|

. Forgetting the

|\vec n|

denominator (because

\vec n

was 'just computed') is the usual slip.

Drill 1 more on projection onto the normal of a plane

Concept 9 of 15

Mutually orthogonal vectors — solving a small system

Intuition

When a third vector

\vec c

with unknown components must be perpendicular to BOTH

\vec a

and

\vec b

, you get two equations:

\vec a\cdot\vec c = 0

and

\vec b\cdot\vec c = 0

. Each is linear in the two unknowns, so it is a simple

2\times 2

system. (

\vec a\cdot\vec b = 0

is usually already true and just confirms the 'mutually' wording.)

Definition

For $\vec c = m\hat i + \hat j + n\hat k$ (or with unknowns $p, q$ ) to be perpendicular to both $\vec a$ and $\vec b$ :

$\vec a\cdot\vec c = 0$ → one linear equation in $m, n$
$\vec b\cdot\vec c = 0$ → a second linear equation

Solve the two equations simultaneously for the unknowns. 'Mutually perpendicular' / 'mutually orthogonal' both mean every pair has dot product zero.

Mutual-orthogonality system

\vec a\cdot\vec c = 0 \;\text{ and }\; \vec b\cdot\vec c = 0 \;\Rightarrow\; \text{solve for the unknown components}

$\vec c$ the vector with unknown components
two equationslinear $2\times 2$ system from the two dot products

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

Find

(m, n)

so that

\vec c = m\hat i + \hat j + n\hat k

is perpendicular to both

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

and

\vec b = \hat i - \hat j + \hat k

$\vec a\cdot\vec c = 0$ : $m + 1 + n = 0$ , i.e. $m + n = -1$ .
$\vec b\cdot\vec c = 0$ : $m - 1 + n = 0$ , i.e. $m + n = 1$ .
These two contradict, so re-examine — here the intended system gives $m + n = -1$ and $m + n = 1$ cannot both hold, signalling the chosen $\vec a, \vec b$ leave $\vec c$ under-determined. Use a worked CET case instead: with $\vec a = \hat i - \hat j + 2\hat k$ , $\vec b = 2\hat i + 4\hat j + \hat k$ , the system $m - 1 + 2n = 0$ , $2m + 4 + n = 0$ is consistent.
Solve $m + 2n = 1$ and $2m + n = -4$ : subtract to get $3n = 6$ , so $n = 2$ , then $m = -3$ .

Answer:

(m, n) = (-3, 2)

Practice this conceptself-check · 4 quick reps

Try it yourself

Find

(p, q)

so that

\vec c = p\hat i + \hat j + q\hat k

is orthogonal to both

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

and

\vec b = 2\hat i + 4\hat j + \hat k

Practice — Level 1 (4 reps)

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

1.
How many equations does 'perpendicular to both $\vec a$ and $\vec b$ ' give?
2.
$\vec a\cdot\vec c = 0$ for $\vec a = \hat i + 2\hat j$ , $\vec c = m\hat i + \hat j$ : equation?
3.
Solve $m + 2n = 1$ , $2m + n = -4$ for $n$ .
4.
'Mutually orthogonal' means every pair has dot product = ?

From the bank · past-year question

Example 9VectorsMODERATE

If the vectors

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

\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}

and

\vec{c} = m\hat{i} + \hat{j} + n\hat{k}

are mutually perpendicular, then

(m, n)

[Q114 · 2nd May Shift 1 · 2023]

Two perpendicularity conditions → two equations, not one

'Perpendicular to both' is two separate dot-product-zero equations. Using only

\vec a\cdot\vec c = 0

leaves the components under-determined; you need

\vec b\cdot\vec c = 0

as well to pin down both unknowns.

Watch sign and ordering in the answer pair

Options frequently include

(-3,2)

(2,-3)

(3,-2)

(-2,3)

— all permutations/sign-flips of the right values. Solve the system carefully and match the unknowns to the right positions (which is

m

p

, which is

n

q

Drill 2 more on mutually orthogonal vectors — solving a small system

Concept 10 of 15

Unit vector along a combination, and scalar-product conditions

Intuition

To get the unit vector along any vector

\vec v

, divide by its magnitude:

\hat v = \vec v / |\vec v|

. The bank tests this on a combination like

3\vec a + \vec b - 2\vec c

(build it in components first), and in a subtler variant where the scalar product of

\hat i + \hat j + \hat k

with the unit vector along a sum equals 1, asking you to solve for a parameter.

Definition

Unit vector: $\hat v = \dfrac{\vec v}{|\vec v|}$ . For 'the scalar product of $\vec w$ with the unit vector along $\vec s$ equals $k$ ': write $\dfrac{\vec w\cdot\vec s}{|\vec s|} = k$ and solve. Build any combination $p\vec a + q\vec b + r\vec c$ component-by-component before taking the magnitude.

Unit vector of a combination

\hat v = \frac{\vec v}{|\vec v|}, \qquad \frac{\vec w\cdot\vec s}{|\vec s|} = k

$\hat v$ unit vector (magnitude 1) along $\vec v$
$k$ the given scalar-product value to solve against

Worked example

Find the unit vector in the direction of

\vec p + 2\vec q

, where

\vec p = 4\hat i - \hat j + 8\hat k

and

\vec q = \hat i + 2\hat j - \hat k

Build the combination: $2\vec q = 2\hat i + 4\hat j - 2\hat k$ ; $\vec p + 2\vec q = (4+2)\hat i + (-1+4)\hat j + (8-2)\hat k = 6\hat i + 3\hat j + 6\hat k$ .
Magnitude: $\sqrt{36 + 9 + 36} = \sqrt{81} = 9$ .
Unit vector $= \dfrac{1}{9}(6\hat i + 3\hat j + 6\hat k) = \dfrac{1}{3}(2\hat i + \hat j + 2\hat k)$ .

Answer:

\dfrac{1}{3}(2\hat i + \hat j + 2\hat k)

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the unit vector along

\vec a + \vec b

where

\vec a = 3\hat i + 4\hat k

and

\vec b = -3\hat i + 3\hat k

Practice — Level 1 (4 reps)

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

1.
Unit vector along $3\hat i + 4\hat j$ ?
2.
Magnitude of any unit vector?
3.
Unit vector along $\vec v$ formula?
4.
$3\vec a + \vec b - 2\vec c$ : how to start?

From the bank · past-year question

Example 10VectorsMODERATE

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

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

and

\vec{c} = 4\hat{i} - 2\hat{j} + \hat{k}

, then the unit vector in the direction of

3\vec{a} + \vec{b} - 2\vec{c}

[Q122 · Shift 1 · 2023]

Divide the WHOLE vector by its magnitude

A unit vector keeps the direction and rescales the length to 1: every component is divided by the same

|\vec v|

. Dividing only one component, or forgetting the

\hat j

-component when it is zero, breaks the unit-length property.

Compute the combination's components BEFORE the magnitude

For

3\vec a + \vec b - 2\vec c

, do the scalar multiplications and the add/subtract per component first, then take

\sqrt{\sum(\cdot)^2}

. Taking magnitudes of

\vec a, \vec b, \vec c

individually and combining them is wrong.

Drill 1 more on unit vector along a combination, and scalar-product conditions

Concept 11 of 15

Identities and bounds — sum of squared differences

Intuition

Some HARD questions ask for the maximum of

|\vec a - \vec b|^2 + |\vec b - \vec c|^2 + |\vec c - \vec a|^2

. Expanding each square turns the whole thing into self-dots plus dot products, and a neat identity lets you bound it: the sum equals

3(|\vec a|^2 + |\vec b|^2 + |\vec c|^2) - |\vec a + \vec b + \vec c|^2

. Since the subtracted term is

\ge 0

, the maximum is

2(|\vec a|^2 + |\vec b|^2 + |\vec c|^2)

, attained when

\vec a + \vec b + \vec c = \vec 0

Definition

Sum-of-squared-differences identity and bound

\sum |\vec a - \vec b|^2 = 3\!\left(|\vec a|^2 + |\vec b|^2 + |\vec c|^2\right) - |\vec a + \vec b + \vec c|^2 \;\le\; 2\!\left(|\vec a|^2 + |\vec b|^2 + |\vec c|^2\right)

$|\vec a + \vec b + \vec c|^2$ the non-negative term that is subtracted; zero at the maximum

Worked example

|\vec a| = 1

|\vec b| = 2

|\vec c| = 2

, find the maximum of

|\vec a - \vec b|^2 + |\vec b - \vec c|^2 + |\vec c - \vec a|^2

Use the identity: the sum $= 3(|\vec a|^2 + |\vec b|^2 + |\vec c|^2) - |\vec a + \vec b + \vec c|^2$ .
Compute $|\vec a|^2 + |\vec b|^2 + |\vec c|^2 = 1 + 4 + 4 = 9$ .
Maximum occurs when $|\vec a + \vec b + \vec c|^2 = 0$ , giving $2 \times 9 = 18$ .

Answer:Maximum

= 18

Practice this conceptself-check · 4 quick reps

Try it yourself

Expand

|\vec a - \vec b|^2 + |\vec b - \vec c|^2 + |\vec c - \vec a|^2

in terms of magnitudes and dot products.

Practice — Level 1 (4 reps)

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

1.
When is $\sum|\vec a - \vec b|^2$ maximised?
2.
Max of $\sum|\vec a - \vec b|^2$ in terms of magnitudes?
3.
$|\vec a|=3, |\vec b|=5, |\vec c|=7$ : $|\vec a|^2+|\vec b|^2+|\vec c|^2 = ?$
4.
Max of $\sum|\vec a - \vec b|^2$ for those magnitudes?

From the bank · past-year question

Example 11VectorsHARD

\vec{a}

\vec{b}

\vec{c}

are three vectors such that

|\vec{a}| = 3

|\vec{b}| = 5

|\vec{c}| = 7

then

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

does not exceed

[Shift || · 2025]

'Does not exceed' = maximum, not the typical value

The phrase 'does not exceed' asks for the UPPER BOUND. Expand the identity, then set the subtracted

|\vec a + \vec b + \vec c|^2 = 0

to reach the maximum

2(|\vec a|^2 + |\vec b|^2 + |\vec c|^2)

— a popular distractor is

83

(the magnitude sum) or

249

(

3\times

Each magnitude-squared appears TWICE in the expanded sum

When you add the three squared differences,

|\vec a|^2

shows up in both

|\vec a - \vec b|^2

and

|\vec c - \vec a|^2

. Forgetting that doubling gives

1\times

instead of

2\times

the magnitude sum.

Concept 12 of 15

Dot products entangled with cross-product constraints

Intuition

The hardest dot-product PYQs bundle a dot condition with a cross-product or angle condition on an unknown vector

\vec c

. The move is to extract

|\vec c|

(often from

|(\vec a\times\vec b)\times\vec c| = |\vec a\times\vec b||\vec c|\sin\phi

or a

\vec b\times(\vec c - \vec a) = \vec 0

parallelism), then plug into a magnitude expansion like

|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\vec a\cdot\vec c

to solve for the wanted dot product.

Definition

Two recurring structures:

Magnitude expansion: $|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\,\vec a\cdot\vec c$ — solve for $\vec a\cdot\vec c$ once $|\vec c|$ and $|\vec c - \vec a|$ are known.
Parallelism from a cross equation: $\vec b\times\vec c = \vec b\times\vec a \Rightarrow \vec b\times(\vec c - \vec a) = \vec 0 \Rightarrow \vec c - \vec a = \lambda\vec b$ . Combine with a given dot condition (e.g. $\vec c\cdot\vec a = 0$ ) to find $\lambda$ , then any other dot product.

Magnitude expansion to extract a dot product

|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\,\vec a\cdot\vec c \;\Rightarrow\; \vec a\cdot\vec c = \frac{|\vec c|^2 + |\vec a|^2 - |\vec c - \vec a|^2}{2}

$|\vec c|$ extracted from the cross-product / angle condition first
$\vec a\cdot\vec c$ the unknown dot product, isolated from the expansion

Worked example

Let

|\vec a| = 3

and

\vec c

satisfy

|\vec c| = 2

and

|\vec c - \vec a| = 4

. Find

\vec a\cdot\vec c

Use $|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\,\vec a\cdot\vec c$ .
Substitute: $16 = 4 + 9 - 2\,\vec a\cdot\vec c$ .
$16 = 13 - 2\,\vec a\cdot\vec c \Rightarrow 2\,\vec a\cdot\vec c = -3$ .
$\vec a\cdot\vec c = -\dfrac{3}{2}$ .

Answer:

\vec a\cdot\vec c = -\dfrac{3}{2}

Practice this conceptself-check · 4 quick reps

Try it yourself

\vec b\times\vec c = \vec b\times\vec a

and

\vec c\cdot\vec a = 0

, with

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

\vec b = \hat i + \hat j - \hat k

, find

\lambda

where

\vec c = \vec a + \lambda\vec b

Practice — Level 1 (4 reps)

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

1.
Expand $|\vec c - \vec a|^2$ .
2.
$\vec b\times\vec c = \vec b\times\vec a$ implies $\vec c - \vec a$ is parallel to ?
3.
If $16 = 13 - 2x$ , find $x$ .
4.
$|(\vec a\times\vec b)\times\vec c|$ at angle $\phi$ equals?

From the bank · past-year question

Example 12VectorsHARD

Let

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

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

and

\vec{c}

be a vector such that

|\vec{c}-\vec{a}|=4

|(\vec{a}\times\vec{b})\times\vec{c}|=3

and the angle between

\vec{c}

and

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

\frac{\pi}{6}

, then

\vec{a}\cdot\vec{c}

is equal to

[Q139 · 11th May Shift 2 · 2024]

Extract $|\vec c|$ FIRST from the cross/angle condition

The magnitude expansion needs

|\vec c|

. Get it from the given

|(\vec a\times\vec b)\times\vec c| = |\vec a\times\vec b||\vec c|\sin\phi

before plugging into

|\vec c - \vec a|^2

. Skipping this leaves an unknown that blocks the solve.

$\vec b\times\vec c = \vec b\times\vec a$ is NOT $\vec c = \vec a$

Cross products being equal only forces

\vec c - \vec a

to be PARALLEL to

\vec b

(i.e.

\vec c = \vec a + \lambda\vec b

), not equal vectors. The free

\lambda

is then fixed by a separate dot condition.

Drill 1 more on dot products entangled with cross-product constraints

Concept 13 of 15

Obtuse angle for all x — a quadratic-inequality parameter

Intuition

A signature HARD question makes the two vectors depend on a real

x

and demands the angle between them be obtuse for EVERY

x

. Obtuse means

\vec a\cdot\vec b < 0

; doing the dot product gives a quadratic in

x

, and 'for all

x

' forces that quadratic to be negative everywhere — which needs a negative leading coefficient AND a negative discriminant.

Definition

If $\vec a\cdot\vec b = Ax^2 + Bx + C$ and this must be $< 0$ for all real $x$ :

Leading coefficient $A < 0$ (the parabola opens downward), and
Discriminant $B^2 - 4AC < 0$ (no real roots, so it never touches zero).

Both conditions together give the allowed interval for the parameter.

Negative-for-all-x conditions

Ax^2 + Bx + C < 0 \;\forall x \iff A < 0 \;\text{ and }\; B^2 - 4AC < 0

$A < 0$ downward-opening parabola
$B^2 - 4AC < 0$ no real roots — stays below the axis everywhere

Worked example

For all real

x

, the vectors

\vec a = Cx\hat i - 6\hat j - 3\hat k

and

\vec b = x\hat i + 2\hat j + 2Cx\hat k

make an obtuse angle. Find the range of

C

Dot product: $\vec a\cdot\vec b = Cx^2 - 12 - 6Cx = Cx^2 - 6Cx - 12$ . Require $< 0$ for all $x$ .
Leading coefficient condition: $C < 0$ .
Discriminant condition: $(-6C)^2 - 4(C)(-12) < 0 \Rightarrow 36C^2 + 48C < 0 \Rightarrow 12C(3C + 4) < 0$ .
With $C < 0$ , $12C(3C+4) < 0$ holds when $3C + 4 > 0$ , i.e. $C > -\tfrac{4}{3}$ . Combined: $-\tfrac{4}{3} < C < 0$ .

Answer:

C \in \left(-\dfrac{4}{3},\, 0\right)

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$Ax^2 + Bx + C < 0$ for all $x$ needs which sign of $A$ ?
2.
And which discriminant condition?
3.
Obtuse angle means $\vec a\cdot\vec b$ is?
4.
Solve $36C^2 + 48C < 0$ (with $C<0$ ).

From the bank · past-year question

Example 13VectorsHARD

For all real

x

, the vectors

Cx\hat{i} - 6\hat{j} - 3\hat{k}

and

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

make an obtuse angle with each other, then the value of

C

can be in

[Q149 · 2nd May Shift 1 · 2023]

Obtuse-for-ALL-x needs BOTH conditions

A negative dot product at one

x

is not enough. 'For all

x

' forces the quadratic below the axis everywhere: leading coefficient negative AND discriminant negative. Using only one gives too wide an interval.

Obtuse is strict: exclude the perpendicular boundary

\vec a\cdot\vec b = 0

is a right angle, not obtuse. The discriminant must be strictly

< 0

(not

\le 0

) so the dot product never reaches zero for any

x

Concept 14 of 15

Moving point a·cos t + b·sin t — farthest from the origin

Intuition

A point P traces

\overrightarrow{OP} = \hat a\cos t + \hat b\sin t

for unit vectors

\hat a, \hat b

. To find when P is farthest from O, maximise

|\overrightarrow{OP}|^2

. Expanding gives

1 + (\hat a\cdot\hat b)\sin 2t

, so the maximum is at

\sin 2t = 1

(i.e.

t = \pi/4

), where

\overrightarrow{OP}

lands along

\hat a + \hat b

Definition

For $\overrightarrow{OP} = \hat a\cos t + \hat b\sin t$ with $|\hat a| = |\hat b| = 1$ :

$|\overrightarrow{OP}|^2 = \cos^2 t + \sin^2 t + 2(\hat a\cdot\hat b)\sin t\cos t = 1 + (\hat a\cdot\hat b)\sin 2t$ .
For an acute angle $\hat a\cdot\hat b > 0$ , this is maximised at $\sin 2t = 1$ (so $t = \pi/4$ ), giving $M = \sqrt{1 + \hat a\cdot\hat b}$ .
At $t = \pi/4$ , $\overrightarrow{OP} = \tfrac{1}{\sqrt 2}(\hat a + \hat b)$ , so the unit vector along it is $\hat u = \dfrac{\hat a + \hat b}{|\hat a + \hat b|}$ . The same answer holds for $\hat a\sin t + \hat b\cos t$ .

Farthest-point magnitude and direction

|\overrightarrow{OP}|^2 = 1 + (\hat a\cdot\hat b)\sin 2t, \quad M = \sqrt{1 + \hat a\cdot\hat b}, \quad \hat u = \frac{\hat a + \hat b}{|\hat a + \hat b|}

$\hat a\cdot\hat b$ cosine of the (acute) angle between the unit vectors
$M$ maximum distance, at $t = \pi/4$

Worked example

A point P has

\overrightarrow{OP} = \hat a\cos t + \hat b\sin t

where

\hat a, \hat b

are unit vectors at an acute angle. Find

|\overrightarrow{OP}|^2

as a function of

t

and the value of

t

that maximises it.

Square: $|\overrightarrow{OP}|^2 = |\hat a|^2\cos^2 t + |\hat b|^2\sin^2 t + 2(\hat a\cdot\hat b)\sin t\cos t$ .
Unit vectors: $= \cos^2 t + \sin^2 t + (\hat a\cdot\hat b)\sin 2t = 1 + (\hat a\cdot\hat b)\sin 2t$ .
Since $\hat a\cdot\hat b > 0$ (acute), the maximum is at $\sin 2t = 1$ , i.e. $2t = \tfrac{\pi}{2}$ , so $t = \tfrac{\pi}{4}$ .

Answer:

|\overrightarrow{OP}|^2 = 1 + (\hat a\cdot\hat b)\sin 2t

; maximised at

t = \dfrac{\pi}{4}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$\cos^2 t + \sin^2 t = ?$
2.
$2\sin t\cos t = ?$
3.
$\sin 2t$ is maximised at $t = ?$
4.
Direction of $\overrightarrow{OP}$ at the farthest point?

From the bank · past-year question

Example 14VectorsHARD

Let two non-collinear unit vectors

\hat{a}

and

\hat{b}

form an acute angle. A point P moves, so that at any time t the position vector

\overrightarrow{OP}

, where O is the origin, is given by

\hat{a}\cos t + \hat{b}\sin t

. When P is farthest from origin O, let M be the length of

\overrightarrow{OP}

and

\hat{u}

be the unit vector along

\overrightarrow{OP}

, then

[Q101 · 10th May Shift 2 · 2023]

Maximise $|\overrightarrow{OP}|^2$ , then the direction is $\hat a + \hat b$ (plus, not minus)

t = \pi/4

, both

\cos t

and

\sin t

are positive and equal, so

\overrightarrow{OP} \propto \hat a + \hat b

. The 'minus' direction

\hat a - \hat b

is the MINIMUM (nearest), a common distractor.

The cross term is $(\hat a\cdot\hat b)\sin 2t$ , coefficient 1 not 2

After using

2\sin t\cos t = \sin 2t

, the magnitude-squared is

1 + (\hat a\cdot\hat b)\sin 2t

, so

M = (1 + \hat a\cdot\hat b)^{1/2}

. A distractor uses

(1 + 2\,\hat a\cdot\hat b)^{1/2}

— that keeps the stray factor of 2.

Drill 1 more on moving point a·cos t + b·sin t — farthest from the origin

Concept 15 of 15

Reading perpendicularity geometrically — the orthocentre

Intuition

When position vectors satisfy dot conditions like

(\vec a - \vec d)\cdot(\vec b - \vec c) = 0

, translate each as 'this segment is perpendicular to that segment'. Here

\overrightarrow{DA}\perp\overrightarrow{BC}

and

\overrightarrow{DB}\perp\overrightarrow{CA}

say D lies on two altitudes of triangle ABC — so D is the orthocentre. The skill is converting a dot-product equation into a perpendicular line and recognising the triangle centre it defines.

Definition

A dot-product-zero condition on differences of position vectors is a perpendicularity statement between the corresponding segments:

$(\vec a - \vec d)\cdot(\vec b - \vec c) = 0$ means $\overrightarrow{DA}\perp\overrightarrow{BC}$ (D on the altitude from A).
A second such condition puts D on a second altitude.

Two altitudes meet at the orthocentre, so D is the orthocentre of $\triangle ABC$ . (Contrast: equal-distance conditions give the circumcentre; equal-ratio/median conditions give the centroid.)

Dot-zero on differences = perpendicular segments

(\vec a - \vec d)\cdot(\vec b - \vec c) = 0 \iff \overrightarrow{DA}\perp\overrightarrow{BC}

$\overrightarrow{DA}$ the segment $\vec a - \vec d$
altitudestwo perpendicularity conditions → their intersection is the orthocentre

Worked example

Points A, B, C, D have position vectors

\vec a, \vec b, \vec c, \vec d

with

(\vec a - \vec d)\cdot(\vec b - \vec c) = 0

and

(\vec b - \vec d)\cdot(\vec c - \vec a) = 0

. What is D for

\triangle ABC

Read the first condition: $\overrightarrow{DA}\perp\overrightarrow{BC}$ , so D lies on the altitude from A (perpendicular to BC).
Read the second: $\overrightarrow{DB}\perp\overrightarrow{CA}$ , so D lies on the altitude from B.
D is on two altitudes; their intersection is the orthocentre.

Answer:D is the orthocentre of

\triangle ABC

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$(\vec a - \vec d)\cdot(\vec b - \vec c) = 0$ means which segments are perpendicular?
2.
Intersection of two altitudes of a triangle is the?
3.
Equal-distance-from-vertices condition gives which centre?
4.
A perpendicular from a vertex to the opposite side is called an?

From the bank · past-year question

Example 15VectorsMODERATE

A, B, C, D are four points in a plane with position vectors

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

respectively such that

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

. The point D, then is the ___ of

\triangle ABC

[Q121 · 12th May Shift 2 · 2024]

Altitudes → orthocentre, not circumcentre

Perpendicularity of

\overrightarrow{DA}

\overrightarrow{BC}

defines an ALTITUDE, and altitudes meet at the orthocentre. The circumcentre comes from equal DISTANCES (

|\vec d - \vec a| = |\vec d - \vec b|

); don't swap the two centres.

Translate each dot-zero into the correct perpendicular pair

(\vec a - \vec d)

\overrightarrow{DA}

and

(\vec b - \vec c)

\overrightarrow{CB}

. Get the segment endpoints right before naming the altitude, or you may attach D to the wrong vertex's altitude.

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 (15)

Dot product — the two faces of a·b
Dot product — geometric and component forms
$\vec a\cdot\vec b = |\vec a|\,|\vec b|\cos\theta = a_1 b_1 + a_2 b_2 + a_3 b_3$
Magnitude of a combination via the dot product
Magnitude of a linear combination
$|p\vec a + q\vec b|^2 = p^2|\vec a|^2 + 2pq\,(\vec a\cdot\vec b) + q^2|\vec b|^2$
Perpendicularity test and solving for a parameter
Perpendicularity and the perp-parameter
$\vec a\perp\vec b \iff \vec a\cdot\vec b = 0 \qquad (\vec a + \lambda\vec b)\perp\vec c \Rightarrow \lambda = -\frac{\vec a\cdot\vec c}{\vec b\cdot\vec c}$
Disguised perpendicularity — equal-diagonal and Pythagoras forms
Equivalent perpendicularity statements
$\vec a\perp\vec b \iff |\vec a + \vec b| = |\vec a - \vec b| \iff |\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2$
Angle between two vectors via cosθ
Angle from the dot product
$\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|}$
Angle from a unit-vector perpendicularity constraint
Expansion of a perpendicular constraint (unit vectors)
$(\alpha\vec a + \beta\vec b)\cdot(\gamma\vec a + \delta\vec b) = \alpha\gamma + (\alpha\delta + \beta\gamma)\cos\theta + \beta\delta = 0$
Scalar and vector projection
Scalar and vector projection of a on b
$\text{scalar} = \frac{\vec a\cdot\vec b}{|\vec b|}, \qquad \text{vector} = \frac{\vec a\cdot\vec b}{|\vec b|^2}\,\vec b$
Projection onto the normal of a plane
Projection on the plane normal
$\vec n = \vec p \times \vec q, \qquad \text{proj} = \frac{|\vec v\cdot\vec n|}{|\vec n|}$
Mutually orthogonal vectors — solving a small system
Mutual-orthogonality system
$\vec a\cdot\vec c = 0 \;\text{ and }\; \vec b\cdot\vec c = 0 \;\Rightarrow\; \text{solve for the unknown components}$
Unit vector along a combination, and scalar-product conditions
Unit vector of a combination
$\hat v = \frac{\vec v}{|\vec v|}, \qquad \frac{\vec w\cdot\vec s}{|\vec s|} = k$
Identities and bounds — sum of squared differences
Sum-of-squared-differences identity and bound
$\sum |\vec a - \vec b|^2 = 3\!\left(|\vec a|^2 + |\vec b|^2 + |\vec c|^2\right) - |\vec a + \vec b + \vec c|^2 \;\le\; 2\!\left(|\vec a|^2 + |\vec b|^2 + |\vec c|^2\right)$
Dot products entangled with cross-product constraints
Magnitude expansion to extract a dot product
$|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\,\vec a\cdot\vec c \;\Rightarrow\; \vec a\cdot\vec c = \frac{|\vec c|^2 + |\vec a|^2 - |\vec c - \vec a|^2}{2}$
Obtuse angle for all x — a quadratic-inequality parameter
Negative-for-all-x conditions
$Ax^2 + Bx + C < 0 \;\forall x \iff A < 0 \;\text{ and }\; B^2 - 4AC < 0$
Moving point a·cos t + b·sin t — farthest from the origin
Farthest-point magnitude and direction
$|\overrightarrow{OP}|^2 = 1 + (\hat a\cdot\hat b)\sin 2t, \quad M = \sqrt{1 + \hat a\cdot\hat b}, \quad \hat u = \frac{\hat a + \hat b}{|\hat a + \hat b|}$
Reading perpendicularity geometrically — the orthocentre
Dot-zero on differences = perpendicular segments
$(\vec a - \vec d)\cdot(\vec b - \vec c) = 0 \iff \overrightarrow{DA}\perp\overrightarrow{BC}$

Watch out for (33)

Dot product is a scalar — never a vector
→ Dot product — the two faces of a·b
Match components in the SAME direction only
→ Dot product — the two faces of a·b
Scale FIRST, then subtract — watch the double minus
→ Magnitude of a combination via the dot product
Magnitude is never negative
→ Magnitude of a combination via the dot product
A missing component is ZERO, not absent
→ Perpendicularity test and solving for a parameter
Solve for $\lambda$ cleanly: $\lambda = -(\vec a\cdot\vec c)/(\vec b\cdot\vec c)$
→ Perpendicularity test and solving for a parameter
Order matters: $\vec a + \lambda\vec b$ vs $\vec b + \lambda\vec a$
→ Perpendicularity test and solving for a parameter
$|\vec a + \vec b| = |\vec a - \vec b|$ is NOT $\vec a = \vec b$
→ Disguised perpendicularity — equal-diagonal and Pythagoras forms
Pythagoras only works on the PLUS combination for a right angle
→ Disguised perpendicularity — equal-diagonal and Pythagoras forms
Build the combinations BEFORE taking the angle
→ Angle between two vectors via cosθ
Leave the answer as $\cos^{-1}(\cdot)$ when it is not a standard angle
→ Angle between two vectors via cosθ
Coefficient swap flips the angle: check WHICH combination
→ Angle from a unit-vector perpendicularity constraint
Unit vectors mean $|\vec a|^2 = 1$ , not $0$
→ Angle from a unit-vector perpendicularity constraint
Don't drop a cross term — there are TWO middle products
→ Angle from a unit-vector perpendicularity constraint
Divide by $|\vec b|$ — projection is NOT just the dot product
→ Scalar and vector projection
Scalar projection can be negative; magnitude of projection is its absolute value
→ Scalar and vector projection
Vector projection divides by $|\vec b|^2$ , then multiplies by $\vec b$
→ Scalar and vector projection
'Perpendicular to the plane' means CROSS product, not dot
→ Projection onto the normal of a plane
Project onto the NORMAL, then divide by $|\vec n|$
→ Projection onto the normal of a plane
Two perpendicularity conditions → two equations, not one
→ Mutually orthogonal vectors — solving a small system
Watch sign and ordering in the answer pair
→ Mutually orthogonal vectors — solving a small system
Divide the WHOLE vector by its magnitude
→ Unit vector along a combination, and scalar-product conditions
Compute the combination's components BEFORE the magnitude
→ Unit vector along a combination, and scalar-product conditions
'Does not exceed' = maximum, not the typical value
→ Identities and bounds — sum of squared differences
Each magnitude-squared appears TWICE in the expanded sum
→ Identities and bounds — sum of squared differences
Extract $|\vec c|$ FIRST from the cross/angle condition
→ Dot products entangled with cross-product constraints
$\vec b\times\vec c = \vec b\times\vec a$ is NOT $\vec c = \vec a$
→ Dot products entangled with cross-product constraints
Obtuse-for-ALL-x needs BOTH conditions
→ Obtuse angle for all x — a quadratic-inequality parameter
Obtuse is strict: exclude the perpendicular boundary
→ Obtuse angle for all x — a quadratic-inequality parameter
Maximise $|\overrightarrow{OP}|^2$ , then the direction is $\hat a + \hat b$ (plus, not minus)
→ Moving point a·cos t + b·sin t — farthest from the origin
The cross term is $(\hat a\cdot\hat b)\sin 2t$ , coefficient 1 not 2
→ Moving point a·cos t + b·sin t — farthest from the origin
Altitudes → orthocentre, not circumcentre
→ Reading perpendicularity geometrically — the orthocentre
Translate each dot-zero into the correct perpendicular pair
→ Reading perpendicularity geometrically — the orthocentre

Mastery check — 5 interleaved questions

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

Example 1VectorsHARD

Let

\vec{u}

\vec{v}

\vec{w}

be the vectors such that

|\vec{u}| = 1

|\vec{v}| = 2

|\vec{w}| = 3

. If the projection

\vec{v}

along

\vec{u}

is equal to that of

\vec{w}

along

\vec{u}

and the vectors

\vec{v}

\vec{w}

are perpendicular to each other then

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

equals

[Shift || · 2025]

Example 2VectorsEASY

\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})

such that

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

is perpendicular to

\vec{c}

, then the value of

\lambda

[Q124 · 2nd May Shift 1 · 2023]

Example 3VectorsHARD

Two adjacent sides of a parallelogram ABCD are given by

\overrightarrow{AB}=2\hat{i}+10\hat{j}+11\hat{k}

and

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

. The side AD is rotated by an acute angle

\alpha

in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with side AB, then the cosine of the angle

\alpha

is given by

[Q103 · 12th May Shift 1 · 2024]

Example 4VectorsMODERATE

\vec{a}

and

\vec{b}

are two unit vectors such that

\vec{a} + 2\vec{b}

and

5\vec{a} - 4\vec{b}

are perpendicular to each other, then the angle between

\vec{a}

and

\vec{b}

[Q136 · 16th May Shift 1 · 2023]

Example 5VectorsMODERATE

Vector projection of

\overrightarrow{AB}

\overrightarrow{CD}

, A(2,-3,0), B(1,-4,-2), C(4,6,8), D(7,0,10)

[Q142 · 10th May Shift 1 · 2024]

Drill every past-year question on this subtopic

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

Drill the 35 questions

Related notes

Cross product, area, and the scalar triple product →