JEE Mains Maths · Vector Algebra

Dot Product: Angles and Projections

The dot product in components: the angle between two vectors, perpendicular, acute and obtuse conditions, projections and components along and across a vector, and a vector in the plane of two others.

All Vector Algebra notes JEE Mains Maths strategy

Why this matters

Twenty-seven PYQs. Every one turns a geometric condition — an angle, a projection, lying in a plane — into a dot product, which in components is one line of arithmetic. Three ideas cover the page.

Concept 1 of 3: Angle between two vectors; perpendicular, acute and obtuse

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

, and in components it is

a_1b_1+a_2b_2+a_3b_3

. So the sign of the dot product tells the kind of angle: zero means perpendicular, positive means acute, negative means obtuse. A condition that must hold 'for all

t

' makes the dot product a quadratic in

t

that keeps one sign, which is a discriminant condition.

Definition

$\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$ .
$\vec a\perp\vec b$ exactly when $\vec a\cdot\vec b=0$ .
Acute: $\vec a\cdot\vec b>0$ ; obtuse: $\vec a\cdot\vec b<0$ (excluding the parallel cases).
$(\vec a+\vec b)\perp(\vec a-\vec b)$ exactly when $|\vec a|=|\vec b|$ .

Angle from components

\cos\theta=\frac{a_1b_1+a_2b_2+a_3b_3}{|\vec a|\,|\vec b|}

Worked example

Find the angle between

(1,2,2)

and

(2,-1,2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q152Moderate

Example 1 · Vector Algebra · Dot Product: Angles and Projections

\lambda > 0

, let

\theta

be the angle between the vectors

\overrightarrow{a}=\widehat{i}+ \lambda\widehat{j}- 3\widehat{k}

and

\overrightarrow{b}= 3\widehat{i}-\widehat{j}+ 2\widehat{k}

. If the vectors

\overrightarrow{a}+\overrightarrow{b}

and

\overrightarrow{a}-\overrightarrow{b}

are mutually perpendicular, then the value of

(14\cos\theta)^{2}

is equal to

A negative dot product includes π

\vec a\cdot\vec b<0

also holds when the vectors point in opposite directions. If the question wants a strictly obtuse angle, exclude the antiparallel case.

Drill 6 more on angle between two vectors; perpendicular, acute and obtuse

Concept 2 of 3: Projections and components along and across a vector

The scalar projection of

\vec a

\vec b

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

, the length of

\vec a

's shadow along

\vec b

, with a sign. The projection vector is that length times the unit vector along

\vec b

, that is

\frac{\vec a\cdot\vec b}{|\vec b|^2}\vec b

. What is left over,

\vec a

minus the projection vector, is the component perpendicular to

\vec b

. In a triangle, the projection of

\overrightarrow{AB}

\overrightarrow{AC}

|AB|\cos A

, with

\cos A

from the cosine rule.

Definition

Scalar projection: $\frac{\vec a\cdot\vec b}{|\vec b|}$ . Projection vector: $\frac{\vec a\cdot\vec b}{|\vec b|^2}\vec b$ .
$\vec a=\vec a_\parallel+\vec a_\perp$ , and $|\vec a|^2=|\vec a_\parallel|^2+|\vec a_\perp|^2$ .
A negative projection means the projection vector points opposite $\vec b$ .

Scalar projection

\text{proj}_{\vec b}\vec a=\frac{\vec a\cdot\vec b}{|\vec b|}

Worked example

Find the projection of

(2,3,6)

(1,2,2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 10 · Q62Moderate

Example 2 · Vector Algebra · Dot Product: Angles and Projections

In a triangle

ABC

, if

|\overrightarrow{BC}| = 8,|\overrightarrow{CA}| = 7

|\overrightarrow{AB}| = 10

, then the projection of the vector

\overrightarrow{AB}

\overrightarrow{AC}

is equal to:

|b| for the length, |b|² for the vector

The scalar projection divides by

|\vec b|

; the projection vector divides by

|\vec b|^2

and then multiplies by

\vec b

. Mixing them scales the answer by

|\vec b|

Drill 11 more on projections and components along and across a vector

Concept 3 of 3: A vector in the plane of two others

A vector in the plane of

\vec a

and

\vec b

can be written

\vec v=x\vec a+y\vec b

. Each further condition — a dot product, a projection, a perpendicularity — is one linear equation in

x

and

y

, and two of them fix

\vec v

. When

\vec v

must also be perpendicular to

\vec c

, its direction is

(\vec a\times\vec b)\times\vec c

, and one more condition fixes the length.

Definition

In the plane of $\vec a,\vec b$ : $\vec v=x\vec a+y\vec b$ .
$\vec v\cdot\vec c=x(\vec a\cdot\vec c)+y(\vec b\cdot\vec c)$ : linear in $x,y$ .
In the plane and perpendicular to $\vec c$ : $\vec v\parallel(\vec a\times\vec b)\times\vec c$ .

In the plane of a and b

\vec v=x\,\vec a+y\,\vec b

Worked example

\vec v

lies in the plane of

(1,0,1)

and

(0,1,1)

, with

\vec v\cdot\hat i=2

and

\vec v\cdot\hat j=3

. Find

|\vec v|^2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q166Moderate

Example 3 · Vector Algebra · Dot Product: Angles and Projections

Let

\overrightarrow{a}=\widehat{i}+\widehat{j}+ 2\widehat{k},\overrightarrow{b}= 2\widehat{i}- 3\widehat{j}+\widehat{k}

and

\overrightarrow{c}=\widehat{i}-\widehat{j}+\widehat{k}

be three given vectors. Let

\overrightarrow{v}

be a vector in the plane of

\overrightarrow{a}

and

\overrightarrow{b}

whose projection on

\overrightarrow{c}

\frac{2}{\sqrt{3}}

. If

\overrightarrow{v}\cdot\widehat{j}= 7

, then

\overrightarrow{v}\cdot (\widehat{i}+\widehat{k})

is equal to:

The direction is not the vector

(\vec a\times\vec b)\times\vec c

fixes only the direction. Its length and sign still come from the remaining condition.

Drill 7 more on a vector in the plane of two others

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 (3)

Angle between two vectors; perpendicular, acute and obtuse
Angle from components
$\cos\theta=\frac{a_1b_1+a_2b_2+a_3b_3}{|\vec a|\,|\vec b|}$
Projections and components along and across a vector
Scalar projection
$\text{proj}_{\vec b}\vec a=\frac{\vec a\cdot\vec b}{|\vec b|}$
A vector in the plane of two others
In the plane of a and b
$\vec v=x\,\vec a+y\,\vec b$

Watch out for (3)

A negative dot product includes π
→ Angle between two vectors; perpendicular, acute and obtuse
|b| for the length, |b|² for the vector
→ Projections and components along and across a vector
The direction is not the vector
→ A vector in the plane of two others

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