Vectors JEE Mains Questions - Practice MCQs with Solutions

Important Concepts of Vector Algebra for JEE Main 2027

Before we start solving Vector Algebra questions for JEE Main 2027, let us revise the important concepts of the chapter. In JEE Main Mathematics, you will often find questions from unit vectors, dot product, cross product, projection of vectors and scalar triple product. A clear understanding of these concepts can help candidates to solve the problems faster and more accurately.

Position Vector

A position vector is a vector representing the position of a point in space with respect to the origin. It is one of the basic concepts in the Vector Algebra and is widely used in geometry-based problems.

Formula:

Position Vector of point P(x, y, z) = xi + yj + zk

Unit Vector

A unit vector is a vector having magnitude equal to 1. It is used to indicate the direction of a vector without considering its magnitude.

Formula:

Unit Vector along vector a = a / |a|

Direction Cosines and Direction Ratios

Direction cosines describe the angles that a vector makes with the coordinate axes, while direction ratios are numbers proportional to the components of the vector.

Formula:

$l^2+m^2+n^2=1$

where l, m, and n are the direction cosines of a vector.

Dot Product (Scalar Product)

The dot product of two vectors produces a scalar quantity. It is often used to find the angle between two vectors and also to find if two vectors are perpendicular to each other.

Formula:

$
\mathrm{a} \cdot \mathrm{~b}=|\mathrm{a}||\mathrm{b}| \cos \theta
$
For vectors $\mathrm{a}=\mathrm{a}_1 \mathrm{i}+\mathrm{a}_2 \mathrm{j}+\mathrm{a}_3 \mathrm{k}$ and $\mathrm{b}=\mathrm{b}_1 \mathrm{i}+\mathrm{b}_2 \mathrm{j}+\mathrm{b}_3 \mathrm{k}$,

$
a \cdot b=a_1 b_1+a_2 b_2+a_3 b_3
$

Projection of a Vector

Projection measures the component of one vector in the direction of another vector. Questions based on projection are frequently asked in JEE Main.

Formula:
Scalar Projection of a on $\mathrm{b}=(\mathrm{a} \cdot \mathrm{b}) /|\mathrm{b}|$
Vector Projection of $a$ on $b=\left[(a \cdot b) /|b|^2\right] b$

Cross Product (Vector Product)

The cross product of two vectors produces a vector perpendicular to both given vectors. It is extensively used in area-related problems.

Formula:

$
|\mathrm{a} \times \mathrm{b}|=|\mathrm{a}||\mathrm{b}| \sin \theta
$
Area of Parallelogram $=|\mathrm{a} \times \mathrm{b}|$

Area of Triangle $=(1 / 2)|\mathrm{a} \times \mathrm{b}|$

Scalar Triple Product

The scalar triple product is useful for determining the volume formed by three vectors and checking whether vectors are coplanar.

$
[a b c]=a \cdot(b \times c)
$

Volume of Parallelepiped $=|\mathrm{a} \cdot(\mathrm{b} \times \mathrm{c})|$
Coplanar Condition: $
a \cdot(b \times c)=0
$

Vector Algebra JEE Main Questions with Solutions

Practice more of the Vectors JEE Main previous year questions, and by doing so, you will become more aware of the question pattern, processes involved, and scoring strategies, which will help you score well in this section. Here we are providing JEE Main previous year questions on vectors for you to practice on.

Question 1: Let $
\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \quad \vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}
$

and a vector $\vec{c}$ be such that $
(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}
$

and $
\vec{a} \cdot \vec{c}=3 .
$ If $
\vec{b} \times \vec{c}=\vec{d},
$ then $
|\vec{a} \cdot \vec{d}|
$

is equal to:
1. 18
2. 12
3. 9
4. 15

Solution:

Given:

$
\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \quad \vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}
$

Calculate

$
\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & 1 \\
3 & 2 & 5
\end{array}\right|=-17 \hat{i}-7 \hat{j}+13 \hat{k}
$

Given:

$
(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}
$

Then

$
(\vec{a}-\vec{c}) \times \vec{b}=\vec{a} \times \vec{b}-\vec{c} \times \vec{b}
$

So

$
\begin{gathered}
\vec{b} \times \vec{c}=(-18 \hat{i}-3 \hat{j}+12 \hat{k})-(-17 \hat{i}-7 \hat{j}+13 \hat{k}) \\
\vec{b} \times \vec{c}=-\hat{i}+4 \hat{j}-\hat{k}
\end{gathered}
$

Let

$
\vec{d}=\vec{b} \times \vec{c}
$

Now

$
\begin{aligned}
\vec{a} \cdot \vec{d}= & (2 \hat{i}-3 \hat{j}+\hat{k}) \cdot(-\hat{i}+4 \hat{j}-\hat{k}) \\
& =-2-12-1=-15
\end{aligned}
$

Therefore

$
|\vec{a} \cdot \vec{d}|=15
$
Hence, the correct answer is option (4).

Question 2: Let $\vec{a}$ and $\vec{b}$ be vectors of the same magnitude such that

$
\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=2+\sqrt{2}
$

Then $
\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}
$

is:
1. $2+\sqrt{2}$
2. $\frac{2+\sqrt{4}}{2}$
3. $\frac{4+2 \sqrt{2}}{2}$
4. $1+\sqrt{2}$

Solution:

$
\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=2+\sqrt{2}
$

Applying componendo and dividendo,

$
\begin{aligned}
\frac{2|\vec{a}+\vec{b}|}{2|\vec{a}-\vec{b}|} & =\frac{(2+\sqrt{2})+1}{(2+\sqrt{2})-1} \\
\Rightarrow|\vec{a}+\vec{b}| & =(1+\sqrt{2})|\vec{a}-\vec{b}|
\end{aligned}
$

Squaring,

$
|\vec{a}+\vec{b}|^2=(3+2 \sqrt{2})|\vec{a}-\vec{b}|^2
$

Since $|\vec{a}|=|\vec{b}|$,

$
\begin{gathered}
2|\vec{a}|^2+2 \vec{a} \cdot \vec{b}=(3+2 \sqrt{2})\left(2|\vec{a}|^2-2 \vec{a} \cdot \vec{b}\right) \\
2|\vec{a}|^2(2+2 \sqrt{2})=2 \vec{a} \cdot \vec{b}(4+2 \sqrt{2}) \\
\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}=\frac{2+2 \sqrt{2}}{4+2 \sqrt{2}}=\frac{1}{2}
\end{gathered}
$

Therefore,

$
\begin{gathered}
\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}=1+\frac{|\vec{b}|^2}{|\vec{a}|^2}+\frac{2 \vec{a} \cdot \vec{b}}{|\vec{a}|^2} \\
=1+1+2\left(\frac{1}{2}\right)=3 \\
3=2+\sqrt{2}
\end{gathered}
$
Hence, the correct answer is option (1).

Question 3: Consider two vectors $
\vec{u}=3 \hat{i}-\hat{j}
$ and $
\vec{v}=2 \hat{i}+\hat{j}-\lambda \hat{k}, \quad \lambda>0 .
$ The angle between them is

$
\cos ^{-1}\left(\frac{5}{2 \sqrt{7}}\right) .
$ Let $
\vec{v}=\vec{v}_1+\vec{v}_2
$

where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$.
Then $
\left|\vec{v}_1\right|^2+\left|\vec{v}_2\right|^2
$

is equal to:
1. $\frac{23}{2}$
2. $\frac{25}{2}$
3. 10
4. 14

Solution:

$
\begin{gathered}
\vec{u} \cdot \vec{v}=|\vec{u}||\vec{v}| \cos \theta \\
6-1=\sqrt{10} \sqrt{5+\lambda^2}\left(\frac{5}{2 \sqrt{7}}\right) \\
5=\frac{5}{2} \sqrt{5+\lambda^2} \\
4=5+\lambda^2 \\
\lambda^2=9 \\
\lambda=3
\end{gathered}
$

Now

$
\vec{v}_1=k \vec{u}
$

Since

$
\begin{gathered}
\vec{v}=\vec{v}_1+\vec{v}_2, \\
\vec{v} \cdot \vec{u}=k|\vec{u}|^2 \\
5=10 k \\
k=\frac{1}{2} \\
\vec{v}_1=\frac{1}{2} \vec{u}=\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j} \\
\left|\vec{v}_1\right|^2=\frac{10}{4} \\
\vec{v}_2=\vec{v}-\vec{v}_1=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k} \\
\left|\vec{v}_2\right|^2=\frac{10}{4}+9
\end{gathered}
$

Thus,

$
\left|\vec{v}_1\right|^2+\left|\vec{v}_2\right|^2=\frac{10}{4}+\frac{10}{4}+9=14
$

Hence, the correct answer is option (4).

Question 4: Let the angle

$
\theta, \quad 0<\theta<\frac{\pi}{2}
$

between two unit vectors $\hat{a}$ and $\hat{b}$ be

$
\sin ^{-1}\left(\frac{6 \sqrt{5}}{9}\right)
$

If

$
\vec{c}=3 \hat{a}+6 \hat{b}+9(\hat{a} \times \hat{b}),
$

then the value of

$
9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})
$

is:
1. 31
2. 29
3. 24
4. 27

Solution:

$
\begin{aligned}
& \vec{c} \cdot \hat{a}=3+6(\hat{a} \cdot \hat{b}) \\
& \vec{c} \cdot \hat{b}=3(\hat{a} \cdot \hat{b})+6
\end{aligned}
$

Therefore,

$
9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})=9+45(\hat{a} \cdot \hat{b})
$

Given

$
\begin{gathered}
\sin \theta=\frac{\sqrt{65}}{9} \\
(\hat{a} \cdot \hat{b})^2=1-\left(\frac{\sqrt{65}}{9}\right)^2=\frac{16}{81}
\end{gathered}
$

Since $0<\theta<\frac{\pi}{2}$,

$
\hat{a} \cdot \hat{b}=\frac{4}{9}
$

Thus

$
9+45\left(\frac{4}{9}\right)=29
$
Hence, the correct answer is option (2).

Question 5: If $\vec{a}$ is a non-zero vector such that its projections on the vectors

$
2 \hat{i}-\hat{j}+2 \hat{k}, \quad \hat{i}+2 \hat{j}-2 \hat{k}
$ and $
\hat{k}
$

are equal, then a unit vector along $\vec{a}$ is:
1. $\frac{1}{\sqrt{155}}(-7 \hat{i}+9 \hat{j}+5 \hat{k})$
2. $\frac{1}{\sqrt{155}}(-7 \hat{i}+9 \hat{j}-5 \hat{k})$
3. $\frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}-5 \hat{k})$
4. $\frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}+5 \hat{k})$

Solution:

$
\frac{\vec{a} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})}{3}=\vec{a} \cdot \hat{k}=\frac{\vec{a} \cdot(\hat{i}+2 \hat{j}-2 \hat{k})}{3}
$

Therefore

$
\vec{a} \cdot(2 \hat{i}-\hat{j}-\hat{k})=0
$

and

$
\vec{a} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})=0
$

Hence

$
\begin{aligned}
& \vec{a} \|(2 \hat{i}-\hat{j}-\hat{k}) \times(\hat{i}+2 \hat{j}-5 \hat{k}) \\
& =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & -1 \\
1 & 2 & -5
\end{array}\right|=7 \hat{i}+9 \hat{j}+5 \hat{k}
\end{aligned}
$

Thus the unit vector is

$
\frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}+5 \hat{k})
$

Hence, the correct answer is option (4).

Question 6:

Let the area of the triangle formed by the lines

$
\begin{aligned}
& \frac{x+2}{1}=\frac{y-1}{1}=\frac{z}{1} \\
& \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}
\end{aligned}
$

and

$
\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}
$

be $A$.
Then $A^2$ is equal to $\_\_\_\_$ .

Solution:

$
\begin{gathered}
P=(-2,1,0), \quad Q=(0,3,2), \quad R=(3,0,1) \\
\overrightarrow{P Q}=2 \hat{i}+2 \hat{j}+2 \hat{k} \\
\overrightarrow{P R}=5 \hat{i}-\hat{j}+\hat{k} \\
A=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\sqrt{56} \\
A^2=56
\end{gathered}
$

Hence, the answer is 56.

Question 7: Let $
\vec{a}=\hat{i}+2 \hat{j}+\hat{k}
$ and

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

Let $\hat{c}$ be a unit vector in the plane of $\vec{a}$ and $\vec{b}$ and perpendicular to $\vec{a}$.
Then $\hat{c}$ is:
1. $\frac{1}{5}(\hat{j}-2 \hat{k})$
2. $\frac{1}{3}(-\hat{i}+\hat{j}-\hat{k})$
3. $\frac{1}{3}(\hat{i}-\hat{j}+\hat{k})$
4. $\frac{1}{\sqrt{2}}(-\hat{i}+\hat{k})$

Solution:

$
\begin{gathered}
\vec{c}=x \vec{a}+y \vec{b} \\
=x(\hat{i}+2 \hat{j}+\hat{k})+y(2 \hat{i}+\hat{j}-\hat{k})
\end{gathered}
$

Since

$
\begin{gathered}
\vec{a} \cdot \vec{c}=0 \\
(\hat{i}+2 \hat{j}+\hat{k}) \cdot \vec{c}=0 \\
(x+2 y)+2(2 x+y)+(x-y)=0 \\
6 x+3 y=0 \\
y=-2 x
\end{gathered}
$

Therefore

$
\begin{gathered}
\vec{c}=x(-3 \hat{i}+3 \hat{k}) \\
|\vec{c}|=3 \sqrt{2}|x|
\end{gathered}
$

For unit vector,

$
\begin{gathered}
3 \sqrt{2}|x|=1 \\
x=\frac{1}{3 \sqrt{2}}
\end{gathered}
$

Hence

$
\hat{c}=\frac{1}{\sqrt{2}}(-\hat{i}+\hat{k})
$

Hence, the correct answer is option (4).

Question 8: Let the three sides of a triangle $A B C$ be given by the vectors

$
2 \hat{i}-\hat{j}+\hat{k}, \quad \hat{i}-3 \hat{j}-5 \hat{k}, \quad 3 \hat{i}-4 \hat{j}-4 \hat{k}
$

Let $G$ be the centroid of triangle $A B C$. Then

$
6\left(|\overrightarrow{A G}|^2+|\overrightarrow{B G}|^2+|\overrightarrow{C G}|^2\right)
$

is equal to $\_\_\_\_$ .

Solution:

The given side vectors satisfy

$
(2 \hat{i}-\hat{j}+\hat{k})+(\hat{i}-3 \hat{j}-5 \hat{k})=3 \hat{i}-4 \hat{j}-4 \hat{k}
$

Hence, we may take

$
\overrightarrow{A B}=2 \hat{i}-\hat{j}+\hat{k}
$

and

$$
\overrightarrow{A C}=3 \hat{i}-4 \hat{j}-4 \hat{k} .
$$

Choose $A$ as the origin.

Then

$
\begin{gathered}
A=(0,0,0) \\
B=(2,-1,1) \\
C=(3,-4,-4)
\end{gathered}
$

The centroid is

$
G=\left(\frac{0+2+3}{3}, \frac{0-1-4}{3}, \frac{0+1-4}{3}\right)=\left(\frac{5}{3},-\frac{5}{3},-1\right)
$

Therefore,

$
\begin{aligned}
\overrightarrow{A G} & =\left(\frac{5}{3},-\frac{5}{3},-1\right) \\
|\overrightarrow{A G}|^2 & =\frac{25}{9}+\frac{25}{9}+1=\frac{59}{9}
\end{aligned}
$

Now

$
\begin{gathered}
\overrightarrow{B G}=\left(\frac{5}{3}-2,-\frac{5}{3}+1,-1-1\right)=\left(-\frac{1}{3},-\frac{2}{3},-2\right) \\
|\overrightarrow{B G}|^2=\frac{1}{9}+\frac{4}{9}+4=\frac{41}{9}
\end{gathered}
$

Similarly,

$
\begin{gathered}
\overrightarrow{C G}=\left(\frac{5}{3}-3,-\frac{5}{3}+4,-1+4\right)=\left(-\frac{4}{3}, \frac{7}{3}, 3\right) \\
|\overrightarrow{C G}|^2=\frac{16}{9}+\frac{49}{9}+9=\frac{146}{9}
\end{gathered}
$

Hence,

$
|\overrightarrow{A G}|^2+|\overrightarrow{B G}|^2+|\overrightarrow{C G}|^2=\frac{59+41+146}{9}=\frac{246}{9}=\frac{82}{3}
$

Therefore,

$
6\left(|\overrightarrow{A G}|^2+|\overrightarrow{B G}|^2+|\overrightarrow{C G}|^2\right)=6\left(\frac{82}{3}\right)=164
$

Answer: 164.

Key benefits of learning vectors for JEE Main:

• The use of vectors makes calculations like distance, angles, and meeting points of lines and planes relatively simple. The shortest distance between them could be easily found using the cross product.

• The direction and the magnitude are represented using vectors. Students will easily learn to think in 3D with the help of vector applications that will help them in solving 3D and geometry-based questions.

• The students learn to solve all the maximum/minimum problems directly. Like finding the maximum area of a triangle or the maximum volume of a parallelepiped using dot and cross product formulas, which would help to save time and labor.

• The use of the vector is to link the geometrical concept with algebra using the application of the dot and cross products. This makes angles, perpendicularity, and the area of a plane more understandable.

• Vectors are repeated in JEE Main several times, and if learned, they will ensure maximum marks. When the students get a grasp of vectors, they feel confident in solving 3D geometry and coordinate geometry.

Best Books for Vector Algebra JEE Main 2027

JEE Main Mathematics Chapterwise Weightage 2027

The knowledge of the chapter-wise weightage for the JEE Main exam would enable students to choose topics smartly and prepare efficiently. Though every chapter is of great importance, a few of them mark higher weights and frequent occurrences.

Preparation Strategy for Vector Algebra in JEE Main 2027

An effective preparation strategy will enable the students to clear their concepts of Vector Algebra and will also enable them to score very good marks in JEE Main 2027. Here is a preparation strategy of Vector Algebra for JEE Main 2027 to help you.

1. Cover all NCERT Examples: The first step to prepare for Vector Algebra in JEE Main 2027 would be to cover all solved examples of NCERT mathematics and all exercise questions of the textbook. Covering NCERT makes clear the basic concepts, and many basic formulas are used to solve problems on advanced-level questions.

2. Learn all the Formulas and Identities: Vector Algebra has several important formulas and identities relating to dot product, cross product, scalar triple product, unit vectors and projections, and students must memorize all the formulas and know where they are applied. Having a separate formula notebook could be very beneficial.

3. Solve all the PYQ's of the last 10 years: Previous Year Questions have been one of the best resources for JEE preparation and for Vector Algebra too. Studying the previous 10 years' questions helps to get an idea of the level of the questions and what type of questions are generally asked in the exam.

4. Solve moderate and advanced problems: After being comfortable with the simple and basic problems, students should start practicing some moderate and advanced problems. Solving advanced questions strengthens their problem-solving skills and helps them develop the ability to solve application-based problems in very little time.

5. Give chapter-wise mock tests: Chapter-wise mock tests help the students get an idea of how well they are preparing and in which topics they are lagging. They are conducted similarly to the original test pattern, so it also helps to improve time management.

6. Revise using short notes weekly: Keeping track of all the formulas and concepts learned needs constant revision. Hence, students must prepare their short notes consisting of important formulas, identities, and some quick tricks, which will help them in revising concepts quickly. The short notes should be revised weekly.