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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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, previous year questions, and chapter-wise weightage to boost your preparation. JEE Main 2026 registration is open from 31 October to 27 November 2025, and Session 1 will held from 21 to 30 January 2026. New updates this year include dark mode, adjustable font size, and zoom options. Lt us begin with the article.
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
$
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.
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:
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) |
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)
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.
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.
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.
On Question asked by student community
Hello murali
No, your son is not eligible for OBC NCL for IIT JEE because you fall in the "creamy layer" occupational category, regardless of your current employment status or family income. Students whose family income is less than Rs. 8 lakhs annually and they are not belong to the "creamy layer".
Note -
Thank You
Hello,
You can fill the JEE Main form even if you are a private candidate
Write the
name of the school/board from where you are appearing as a private candidate
.
If your Class 12 admit card or registration slip shows a school/centre name, use that exactly.
If your board lists you as a “Private Candidate” under the board name , then write:
CBSE – Private Candidate
(or your board name – Private Candidate)
Use the pin code of the examination centre/school mentioned on your Class 12 private candidate admit card or registration details.
If your board does not give any school address and only shows the regional office address, then use the regional office address pin code given by your board.
Hope it helps !
Hello Aspirant
You should not leave the OBC-NCL certificate ID blank in the JEE Main form it can create problems later.
NTA wants the certificate details while filling the form, not just at counselling. If you can, apply for the OBC-NCL certificate immediately so you get the ID on time.
If you fail to submit the certificate during counselling, your category will shift to General. It’s safer to enter OBC-NCL only if you’re sure you’ll get the certificate before counselling.
Hope it will help you
Hello,
No, X-rays are not deleted for the JEE 2026 exams. X-rays are part of the electromagnetic spectrum, a topic that remains in the syllabus for both JEE Main and JEE Advanced. Here in this article you will find more about the JEE Main syllabus.
I hope it will clear your query!!
Of course Lakshmi, that's why we are here. As you mentioned, you didn't start preparation for JEE Mains Exam. Don't worry start TODAY. Because today is the best day to start.
Okay you can start with the previous years question papers which you can download by Clicking Here .
You also can browse the JEE Mains Syllabus by Clicking Here . These help you to know which will come in the exam.
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