JEE Mains Maths · Vector Algebra

Vectors in Geometry and Rotation

Vectors applied to points and figures: dividing a segment, collinear points, the centroid, orthocentre and circumcentre, angle bisectors, and rotating a vector or the axes.

All Vector Algebra notes JEE Mains Maths strategy

Why this matters

Twenty-four PYQs. They are geometry questions in vector language: find a point, a length or a ratio. Each rests on one fact — the section formula, the bisector's direction, or that rotation keeps length. Three ideas cover the page.

Concept 1 of 3: Section points, centroid and collinearity

The point dividing

AB

internally in the ratio

m:n

\frac{n\vec a+m\vec b}{m+n}

: the weight on each end is the other part of the ratio. The centroid averages the three vertices. Three points are collinear when

\overrightarrow{AB}

and

\overrightarrow{AC}

are parallel, that is, their components are proportional.

Definition

Internal $m:n$ : $\frac{n\vec a+m\vec b}{m+n}$ . External: $\frac{m\vec b-n\vec a}{m-n}$ .
Centroid $G=\frac{\vec a+\vec b+\vec c}{3}$ ; $|AG|^2+|BG|^2+|CG|^2=\frac13(a^2+b^2+c^2)$ .
Collinear: $\overrightarrow{AB}=t\,\overrightarrow{AC}$ .

Section formula

\vec p=\frac{n\,\vec a+m\,\vec b}{m+n}

Worked example

Find the point dividing

A(1,2,3)

B(4,8,9)

internally in the ratio

1:2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q71Moderate

Example 1 · Vector Algebra · Vectors in Geometry and Rotation

If the points with vectors

\alpha\widehat{i}+ 10\widehat{j}+ 13\widehat{k}

6\widehat{i}+ 11\widehat{j}+ 11\widehat{k},\frac{9}{2}\widehat{i}+ \beta\widehat{j}- 8\widehat{k}

are collinear, then

(19\alpha - 6\beta)^{2}

is equal to

Cross the weights

For

AP:PB=m:n

\vec a

carries weight

n

and

\vec b

weight

m

. Putting

m

\vec a

gives the point dividing in

n:m

Drill 8 more on section points, centroid and collinearity

Concept 2 of 3: Triangle centres and angle bisectors

The internal bisector of the angle between

\vec a

and

\vec b

runs along

\hat a+\hat b

, the sum of the unit vectors, and it meets the opposite side dividing it in the ratio of the adjacent sides. A point equidistant from two lines through a vertex lies on a bisector. With the circumcentre

O

at the origin, the orthocentre is

\vec a+\vec b+\vec c

, and the centroid divides

OH

in the ratio

1:2

Definition

Internal bisector direction: $\hat a+\hat b$ ; external: $\hat a-\hat b$ .
The bisector from $O$ meets $AB$ at $C$ with $AC:CB=OA:OB$ .
Circumcentre at the origin: $H=\vec a+\vec b+\vec c$ , $G=\frac{H}{3}$ ; in general $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OH}$ .
Equilateral triangle: all centres coincide.

Internal bisector

\text{direction}=\frac{\vec a}{|\vec a|}+\frac{\vec b}{|\vec b|}

Worked example

Find the direction of the internal bisector of the angle between

(3,0,0)

and

(0,4,0)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q73Moderate

Example 2 · Vector Algebra · Vectors in Geometry and Rotation

Let

O

be the origin and the position vector of

A

and

B

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

and

2\widehat{i}+ 4\widehat{j}+ 4\widehat{k}

respectively. If the internal bisector of

\angle AOB

meets the line

AB

C

, then the length of

OC

Add unit vectors

\vec a+\vec b

bisects the angle only when

|\vec a|=|\vec b|

. Normalise first:

\hat a+\hat b

Drill 7 more on triangle centres and angle bisectors

Concept 3 of 3: Rotating a vector, and rotating the axes

A rotation keeps length. In the plane, turning

(x,y)

counterclockwise by

\theta

gives

(x\cos\theta-y\sin\theta,\ x\sin\theta+y\cos\theta)

. In space, a vector rotated in the plane of

\vec a

and another vector is a combination of the two, fixed by keeping the length and the angle. Rotating the axes instead leaves the vector alone and changes its components, but

x^2+y^2

stays the same.

Definition

$|\text{rotated vector}|=|\text{original}|$ .
Plane rotation by $\theta$ : $(x\cos\theta-y\sin\theta,\ x\sin\theta+y\cos\theta)$ .
A point at angle $\theta$ from $\overrightarrow{OA}$ on a unit circle: $\cos\theta\,\hat u+\sin\theta\,\hat w$ , $\hat w\perp\hat u$ in the plane.
Rotated axes: new components satisfy $x'^2+y'^2=x^2+y^2$ .

Plane rotation

(x,y)\mapsto(x\cos\theta-y\sin\theta,\ x\sin\theta+y\cos\theta)

Worked example

Rotate

(1,0)

counterclockwise by

60^\circ

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 5 · Q62Moderate

Example 3 · Vector Algebra · Vectors in Geometry and Rotation

Let a vector

\alpha\widehat{i}+ \beta\widehat{j}

be obtained by rotating the vector

\sqrt{3}\widehat{i}+\widehat{j}

by an angle

45^{\circ}

about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices

(\alpha,\beta),(0,\beta)

and

(0,0)

is equal to

Which way it turns

A right-angle rotation has two possible results,

\pm

a perpendicular vector. The question's words — counterclockwise, 'passing through the y-axis' — pick one; check it against them.

Drill 6 more on rotating a vector, and rotating the axes

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)

Section points, centroid and collinearity
Section formula
$\vec p=\frac{n\,\vec a+m\,\vec b}{m+n}$
Triangle centres and angle bisectors
Internal bisector
$\text{direction}=\frac{\vec a}{|\vec a|}+\frac{\vec b}{|\vec b|}$
Rotating a vector, and rotating the axes
Plane rotation
$(x,y)\mapsto(x\cos\theta-y\sin\theta,\ x\sin\theta+y\cos\theta)$

Watch out for (3)

Cross the weights
→ Section points, centroid and collinearity
Add unit vectors
→ Triangle centres and angle bisectors
Which way it turns
→ Rotating a vector, and rotating the axes

Test yourself on Vector Algebra

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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