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Vectors JEE Mains Questions - Practice MCQs with Solutions
Vectors JEE Main 2026 Questions and Solutions: Vector algebra is a crucial topic for JEE Main as it lays the foundation for solving advanced and complex problems in Mathematics. With vectors, we can calculate area and volume, and better understand concepts of direction, angles, and 3D geometry. This makes solving advanced problems much easier. In JEE Main exam, questions from Vector Algebra are often asked directly. If you have studied this chapter thoroughly and practiced its questions, you will be able to solve them easily in the exam.
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JEE Main 2026 is expected to be held in two sessions. NTA holds the entrance exam in two sessions to provide the following benefits:
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- A candidate is not required to take both the sessions, however, if someone appears in both the sessions, then their best NTA scores will be considered for merit list preparation.
This Story also Contains
- Mathematics: Vector PYQs of JEE Main
- Key Benefits of Studying Vectors for JEE Main
- JEE Main Previous Year Mathematics Questions
- JEE Main 2026 Exam Pattern
- JEE Main Mathematics Chapterwise Weightage
In this article, we have provided Vectors questions along with previous year questions for practice. We have also included the exam pattern and the chapter-wise weightage of Mathematics to give you a clear strategy for your studies. Practice Vectors MCQs for JEE Main 2026 with step-by-step solutions, previous year questions, and chapter-wise weightage to boost your preparation
Mathematics: Vector PYQs of JEE Main
By practicing previous year questions, you get a clear understanding of the question pattern, the techniques to solve them, and the scoring strategies that can help you perform better in the exam. Below, we have provided a set of Vectors JEE Main previous year questions for your practice.
Question 1: Let $\vec{a}=2 \hat{i}-3 \hat{j}+ \hat k, \vec{b}=3 \hat{i}+2 \hat{j}+5 \hat k$ 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 $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3$. If $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{d}}$, then $|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}|$ 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 $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & 1 \\
3 & 2 & 5
\end{vmatrix} = -17\hat{i} - 7\hat{j} + 13\hat{k}$
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:
$\vec{b} \times \vec{c} = -18\hat{i} - 3\hat{j} + 12\hat{k} - (-17\hat{i} - 7\hat{j} + 13\hat{k})$
$\vec{b} \times \vec{c} = -\hat{i} + 4\hat{j} - \hat{k}$
Let $\vec{d} = \vec{b} \times \vec{c}$
Now:
$\vec{a} \cdot \vec{d} = (2\hat{i} - 3\hat{j} + \hat{k}) \cdot (-\hat{i} + 4\hat{j} - \hat{k})$
$= -2 - 12 - 1 = -15$
Therefore:
$|\vec{a} \cdot \vec{d}| = 15$
Hence, the answer is option 4.
Question 2: Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is :
1) $2+\sqrt{2}$
2) $2+4 \sqrt{2}$
3) $4+2 \sqrt{2}$
4) $1+\sqrt{2}$
Solution:
$\frac{|\bar{a}+\bar{b}|+|\bar{a}-\bar{b}|}{|\bar{a}+\bar{b}|-|\bar{a}-\bar{b}|}=\sqrt{2}+1$
Apply componendo and dividendo
$\begin{aligned} & \Rightarrow \frac{2|\bar{a}+\bar{b}|}{2|\bar{a}-\bar{b}|}=\frac{\sqrt{2}+2}{\sqrt{2}} \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|=(1+\sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}| \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2=(3+2 \sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}|^2 \\ & \Rightarrow 2|\overline{\mathrm{a}}|^2+2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=(3+2 \sqrt{2})\left(2|\overline{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}\right) \\ & \Rightarrow 2|\overline{\mathrm{a}}|^2(2+2 \sqrt{2})=2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}(4+2 \sqrt{2}) \\ & \Rightarrow \frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}|^2}=\frac{2+2 \sqrt{2}}{4+2 \sqrt{2}}=\frac{1}{\sqrt{2}}\end{aligned}$
Now
$\begin{aligned}
& \frac{|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}=1+\frac{|\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}+\frac{2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{a}}|^2} \\
& =1+1+2\left(\frac{1}{\sqrt{2}}\right)=2+\sqrt{2}
\end{aligned}$
Hence, the correct answer is option 1
Question 3: Consider two vectors $\vec{u}=3 \hat{i}-\hat{j}$ and $\vec{v}=2 \hat{i}+\hat{j}-\lambda \hat{k}, \lambda>0$. The angle between them is given by $\cos ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)$. Let $\vec{v}=\vec{v}_1+\vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\overrightarrow{v_2}$ is perpendicular to $\vec{u}$. Then the value $\left|\overrightarrow{v_1}\right|^2+\left|\overrightarrow{v_2}\right|^2$ is equal to
1) $\frac{23}{2}$
2) $\frac{25}{2}$
3) 10
4) 14
Solution:
$\begin{aligned} & \vec{u} \cdot \vec{v}=|u| \cdot|v| \cdot \cos \theta \\ & \Rightarrow 6-1=\sqrt{10} \cdot \sqrt{5+\lambda^2} \cdot \frac{\sqrt{5}}{2 \sqrt{7}} \\ & \Rightarrow 1=\sqrt{2} \cdot \sqrt{5+\lambda^2} \cdot \frac{1}{2 \sqrt{7}} \\ & \Rightarrow 14=5+\lambda^2 \\ & \Rightarrow \lambda^2=9 \\ & \Rightarrow \lambda=3\end{aligned}$
$\begin{aligned} & v_1=k \vec{u} \\ & \vec{v}=\vec{v}_1+\vec{v}_2 \\ & \Rightarrow \vec{v}=k \vec{u}+\vec{v}_2 \\ & \vec{v} \cdot \vec{u}=k \cdot|\vec{u}|^2\end{aligned}$
$\begin{aligned} & \Rightarrow 5=k \cdot 10 \Rightarrow k=\frac{1}{2} \\ & \therefore \quad \vec{v}_1=\frac{\vec{u}}{2}=\frac{3 \hat{i}}{2}-\frac{\hat{j}}{2}\end{aligned}$
$\begin{aligned} & \left|\vec{v}_1\right|^2=\frac{10}{4} \\ & \vec{v}_2=\vec{v}-\vec{v}_1\end{aligned}$
$=\frac{1}{2} \hat{i}+\frac{3 \hat{j}}{2}-3 \hat{k}$
$\begin{aligned} & \left|\vec{v}_2\right|^2=\frac{10}{4}+9 \\ & \left|\vec{v}_1\right|^2+\left|\vec{v}_2\right|^2=\frac{10}{4}+\frac{10}{4}+9=14\end{aligned}$
Hence, the correct answer is option 4
Question 4: Let the angle $\theta, 0<\theta<\frac{\pi}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\vec{c}=3 \hat{a}+6 \hat{b}+9(\hat{a} \times \hat{b})$, 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:
To solve the problem, we begin with
$\vec{c}=3 \hat{a}+6 \hat{b}+9(\hat{a} \times \hat{b})$
Let's calculate:
$\begin{aligned}
& \vec{c} \cdot \hat{a}=3+6(\hat{a} \cdot \hat{b}) \\
& \vec{c} \cdot \hat{b}=3(\hat{a} \cdot \hat{b})+6
\end{aligned}$
We need to find the value of $9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})$.
Substituting the expressions, we get:
$\begin{aligned} & 9(\vec{c} \cdot \hat{a})=9(3+6 \hat{a} \cdot \hat{b})=27+54(\hat{a} \cdot \hat{b}) \\ & 3(\vec{c} \cdot \hat{b})=3(3 \hat{a} \cdot \hat{b}+6)=9 \hat{a} \cdot \hat{b}+18\end{aligned}$
Therefore,
$9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})=(27+54(\hat{a} \cdot \hat{b}))-(9 \hat{a} \cdot \hat{b}+18)$
Simplifying,
$\begin{aligned}
& =27-18+54(\hat{a} \cdot \hat{b})-9(\hat{a} \cdot \hat{b}) \\
& =9+45(\hat{a} \cdot \hat{b})
\end{aligned}$
Given $\sin \theta=\frac{\sqrt{65}}{9}$, and knowing that for unit vectors $\cos \theta=\hat{a} \cdot \hat{b}$, we use the identity $(\cos \theta)^2=1-(\sin \theta)^2$ :
$\begin{aligned}
& (\hat{a} \cdot \hat{b})^2=1-\left(\frac{\sqrt{65}}{9}\right)^2 \\
& =1-\frac{65}{81} \\
& =\frac{16}{81}
\end{aligned}$
Thus, $\hat{a} \cdot \hat{b}=\frac{4}{9}$.
Substitute back into the equation:
$9+45 \times \frac{4}{9}=9+20=29$
Hence, the correct answer is option 2
Question 5: If $\overrightarrow{\mathrm{a}}$ is a nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k}, \hat{i}+2 \hat{j}-2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\overrightarrow{\mathrm{a}}$ 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:
Projection of $\vec{a}$ on $\vec{v}$
$=\frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}-$
$\begin{aligned} & \Rightarrow \frac{\vec{a} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})}{3}=\frac{\vec{a} \cdot \hat{k}}{1}=\frac{\vec{a} \cdot(\hat{i}+2 \hat{j}-2 \hat{k})}{3} \\ & \Rightarrow \vec{a} \cdot(2 \hat{i}-\hat{j}-\hat{k})=0 \text { and } \vec{a} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})=0 \\ & \Rightarrow \vec{a} \perp(2 \hat{i}-\hat{j}-\hat{k}) \text { and }(\hat{i}+2 \hat{j}-5 \hat{k}) \\ & \Rightarrow \vec{a} \|(2 \hat{i}-\hat{j}-\hat{k}) \times(\hat{i}+2 \hat{j}-5 \hat{k})\end{aligned}$
$\begin{aligned} & \Rightarrow \vec{a}= \pm k\left|\begin{array}{ccc}\hat{i} & -\hat{j} & \hat{k} \\ 2 & -1 & -1 \\ 1 & 2 & -5\end{array}\right|= \pm k(7 \hat{i}+9 \hat{j}-5 \hat{k}) \\ & \Rightarrow \text { Unit vector will be } \frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}+5 \hat{k})\end{aligned}$
Hence, the correct answer is option 4
Question 6: Let the area of the triangle formed by the lines $x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to ______
Solution:
$L_1=\frac{x+2}{1}=\frac{y-1}{1}=\frac{z}{1}=\lambda$, any point on it $(\lambda-2, \lambda$ $+1, \lambda)$
$L_2=\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}=\mu$, any point on it $(5 \mu+3$, $-\mu, \mu+1)$
$L_3=\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}=k$, any point on it $(-3 k$, $3 k+3, k+2)$
$P \equiv$ point of intersection of $L_1$ and $L_2=(-2,1,0)$
$Q=$ point of intersection of $L_1$ and $L_3=(0,3,2)$
$R \equiv$ point of intersection of $L_2$ and $L_3=(3,0,1)$
$\begin{aligned} & \overline{P Q}=2 \hat{i}+2 \hat{j}+2 \hat{k} \\ & \overline{P R}=5 \hat{i}-\hat{j}+\hat{k}\end{aligned}$
$A=\frac{1}{2}|\overline{P Q} \times \overline{P R}|=\sqrt{56}$
$A^2=56$
Hence, the answer is (56).
Question 7:Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$. Let $\hat{\mathrm{c}}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{\mathbf{c}}$ is :
1) $\frac{1}{\sqrt{5}}(\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$
2) $\frac{1}{\sqrt{3}}(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$
3) $\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$
4) $\frac{1}{\sqrt{2}}(-\hat{\mathrm{i}}+\hat{\mathrm{k}})$
Solution:
$\begin{aligned} & \vec{c}=x \vec{a}+y \vec{b} \\ & \vec{c}=x(\hat{i}+2 \hat{j}+\hat{k})+y(2 \hat{i}+\hat{j}-\hat{k}) \\ & \vec{a} \cdot \vec{c}=(\hat{i}+2 \hat{j}+\hat{k}) \cdot(x(\hat{i}+2 \hat{j}+\hat{k})+y(2 \hat{i}+\hat{j}-\hat{k})) \\ & (\hat{i}+2 \hat{j}+\hat{k}) \cdot(x \hat{i}+2 x \hat{j}+x \hat{k})+2 y \hat{i}+y \hat{j}-y \hat{k}=0 \\ & \Rightarrow \quad(x+2 y)+2(x+9)+(x-y)=0 \\ & \Rightarrow \quad y=-2 x \\ & \therefore \quad \vec{c}=x(-3 \hat{i}+3 \hat{k}) \\ & |\vec{c}|=|x| \sqrt{9+9}=3|x| \sqrt{2} \\ & \therefore|\vec{c}|=1 \\ & 3|x| \sqrt{2}=1 \\ & |x|=\frac{1}{3 \sqrt{2}} \\ & \text { Let } x=\frac{1}{3 \sqrt{2}} \\ & \vec{c}=\frac{1}{3 \sqrt{2}}(-3 \hat{i}+3 \hat{k}) \\ & \text { or } \vec{c}=\frac{1}{\sqrt{2}}(-\hat{i}+\hat{k})\end{aligned}$
Hence, the answer is option (4).
Question 8: Let the three sides of a triangle $A B C$ be given by the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$. Let G be the centroid of the triangle ABC . Then $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right)$ is equal to ______
Solution:
The position vector of a point is usually taken from the origin. If a point $ A $ is at the origin, then $ \vec{A} = \vec{0} $.
For two points $ A $ and $ B $, the vector $ \vec{AB} $ is given by:
$
\vec{AB} = \vec{B} - \vec{A}
$
The centroid $ G $ of a triangle with vertices having position vectors $ \vec{A}, \vec{B}, \vec{C} $ is:
$
\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}
$
The vector from point $ A $ to centroid $ G $ is:
$
\vec{AG} = \vec{G} - \vec{A}
$
The square of the magnitude (length) of a vector $ \vec{v} = a\hat{i} + b\hat{j} + c\hat{k} $ is:
$
|\vec{v}|^2 = a^2 + b^2 + c^2
$
Key Benefits of Studying Vectors for JEE Main
Simplifies 3D Geometry Problems – Vectors are useful to make it simple to calculate distances, angles, and intersections of points, lines, and planes. Suppose the shortest distance between skew lines can be easily and effortlessly found by utilizing the cross product.
Boosts Visualization Skills – As we know vectors deal with both magnitude and direction which helps students to think in three dimensions, which is crucial for solving 3D questions and helps in solving geometry-based problems.
Useful for Maximum/Minimum Problems – When you optimize questions like the maximum area of a triangle or the volume of a parallelepiped, it can be solved directly with the use of dot and cross product formulas, which saves time and effort.
Bridges Algebra and Geometry – Dot products and cross products connect algebraic calculations with geometric interpretation, which helps to explain angles, perpendicularity, and areas easily.
Continuously Tested Topic – Vectors come repeatedly, which makes them a high-scoring topic. When you master it, it ensures better accuracy and confidence in 3D geometry and coordinate geometry problems.
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JEE Main Previous Year Mathematics Questions
Practicing previous year questions is one of the best ways to understand the exam pattern, difficulty level, and frequently asked concepts in JEE Main Mathematics. We have provided a collection of solved questions along with Vector Algebra questions from past year papers to help you prepare effectively. Download the eBook given below and start practicing:
JEE Main 2026 Exam Pattern
Knowing the exam pattern is important to plan your preparation and manage time effectively during the test. The pattern of the exam, marking scheme, and type of questions asked in JEE Main 2026 are given below:
|
Particulars |
Details |
|
Mode of Exam |
Computer-Based Test (CBT) |
|
Subjects |
Physics, Chemistry, Mathematics |
|
Total Questions |
75 (25 per subject: 20 MCQs + 5 Numerical Value Questions) |
|
Questions to Attempt |
All questions are compulsory |
|
Type of Questions |
Multiple Choice Questions (MCQs) + Numerical Value Questions |
|
Maximum Marks |
300 |
|
Marking Scheme |
+4 for correct answer -1 for incorrect MCQ No negative marking for numerical questions |
|
Duration |
3 Hours |
|
Medium of Exam |
English, Hindi, Gujarati, and other regional languages (as opted) |
JEE Main Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters
JEE Main Mathematics Chapterwise Weightage
Understanding the chapter-wise weightage for JEE Main Mathematics 2026 helps you prioritize topics and plan your preparation effectively. While every chapter is important, some chapters carry higher marks and appear more frequently in exams
|
Chapter Name | Approximate Weightage (%) |
|
Co-ordinate geometry |
17.89% |
|
Integral Calculus |
10.74% |
|
Limit, continuity, and differentiability |
8.84% |
|
Sets, Relations, and Functions |
7.79% |
|
Complex numbers and quadratic equations |
6.95% |
|
Sequence and series |
7.37% |
|
Matrices and Determinants |
7.16% |
|
Statistics and Probability |
6.32% |
|
Binomial theorem and its simple applications |
4.21% |
|
Three-Dimensional Geometry |
5.26% |
|
Trigonometry |
4.42% |
|
Vector Algebra |
4.84% |
|
Permutations and combinations |
4.21% |
|
Differential equations |
4.00% |
Also Read: JEE Main Mathematics Syllabus 2026
Frequently Asked Questions (FAQs)
Q: Are vectors jee mains questions simple enough to get a good score in JEE Main?
A:
Yes. Most vector questions include formulas and concepts. If you regularly do your practice, you can solve them fastly, and it will make vectors a high-yield topic for getting good marks.
Q: What are Vectors in Mathematics? Why are Vectors Important during the JEE Main exam?
A:
Vectors are quantities that include both magnitude and direction, such as displacement, velocity, or force. They are important in JEE Main as many 3D geometry and coordinate geometry problems are resolved easily if we utilize vector methods.
Q: Do vectors only appear in mathematics?
A:
No. Vectors are also used majorly in physics, like in problems involving motion, forces, and equilibrium. If you study vectors in mathematics, your understanding and problem-solving skills of physics will be strengthened.
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Hello
Here is the short guidance for you to score well and prepare for JEE
-
Stick to a smart routine – Prioritize high-weightage topics and revise daily with a focused timetable.
-
Solve PYQs & mock tests – Practice like it’s the real thing; analyze mistakes without panic.
-
Quality over quantity – Understand concepts deeply rather than rushing through chapters.
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Stay healthy & rested – Your brain needs sleep, water, and breaks to perform at its best.
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Trust your journey – Stay calm, believe in your preparation, and avoid comparing with others.
Hello
Since you’re reappearing for your Class 12 boards in 2026, that will be counted as your official passing year. So, you will be eligible to write JEE Advanced in 2026 and 2027. Although you submitted the boards in 2025, that attempt won’t be considered due to the compartment. JEE Advanced rules allow 2 attempts in 2 consecutive years after passing 12th. So you still have both chances left, which is great! Just make sure you meet the other eligibility conditions too. Keep your focus strong, you’ve got this!
Hello aspirant,
Like all competitive exams, JEE MAINS requires careful preparation and strategy to pass. The JEE Main exam is extremely competitive, so it's critical to carefully plan and strategy. Students can increase their speed, accuracy, and confidence by adhering to a systematic study schedule and practicing frequently.
To know complete preparation tips, you can visit our site through following link:
https://engineering.careers360.com/articles/jee-main-preparation-tips
Thank you
Hello aspirant,
Start with NCERT in every subject for JEE Main. For physics, use H.C. Verma and D.C. Pandey; for chemistry, use R.C. Mukherjee and O.P. Tandon; and for mathematics, use R.D. Sharma or Cengage. J.D. Lee assists with inorganic chemistry, while Morrison & Boyd assists with organic chemistry. Practice chapter-by-chapter Arihant or MTG past year's papers. For efficient preparation, concentrate on clear concepts, frequent revision, and passing practice exams.
Thank you
Hello,
Generally an income certificate isn't required for the JEE Main registration, but if you want to claim the EWS quota, then you need this. You must provide the certificate, issued by a government authority, as proof of your family's income being below the specified limit for the reservation category you wish to apply under.
I hope it will clear your query!!
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