Vector Algebra JEE Main previous year questions with solutions

5 solved JEE Main questions on Vector Algebra, free to read — no sign-in needed. The full chapter has 272 questions; sign in to attempt the remaining 267 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Algebra of Vectors
   Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is the point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is the point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to :
   1. A.$\dfrac{1}{5}(\vec{a} - 4\vec{b})$
   2. B.$\dfrac{1}{5}(\vec{b} - 4\vec{a})$
   3. C.$\dfrac{1}{5}(-\vec{a} + 4\vec{b})$
   4. D.$\dfrac{1}{5}(-\vec{b} + 4\vec{a})$
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   Answer: (B)

   Given $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. Since $\vec{OP} = 5\vec{OR}$, we have $\vec{OR} = \dfrac{\vec{a}}{5}$. Also, $\vec{OQ} = 5\vec{RM}$, which gives $\vec{RM} = \dfrac{\vec{b}}{5}$. The position vector of $M$ is given by $\vec{OM} = \vec{OR} + \vec{RM} = \dfrac{\vec{a}}{5} + \dfrac{\vec{b}}{5}$. Now, we need to find $\vec{PM}$: $\vec{PM} = \vec{OM} - \vec{OP}$ $\vec{PM} = \left(\dfrac{\vec{a}}{5} + \dfrac{\vec{b}}{5}\right) - \vec{a}$ $\vec{PM} = \dfrac{\vec{b} - 4\vec{a}}{5} = \dfrac{1}{5}(\vec{b} - 4\vec{a})$ Answer: $\dfrac{1}{5}(\vec{b} - 4\vec{a})$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Product of 2 vectors
   <p>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 the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is :</p>
   1. A.$\frac{1}{\sqrt{5}}(\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$
   2. B.<p>$\frac{1}{\sqrt{3}}(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$</p>
   3. C.$\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$
   4. D.<p>$\frac{1}{\sqrt{2}}(-\hat{\mathrm{i}}+\hat{\mathrm{k}})$</p>
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   Answer: (D)

   <p>Let vector $\vec{p}$ in plane of $\vec{a} \& \vec{b}=K(\vec{a}+\lambda \vec{b})$ $\begin{aligned} & \overrightarrow{\mathrm{p}} \perp \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow \mathrm{~K}(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}=0 \\ & \Rightarrow 6+\lambda(3)=0 \\ & \Rightarrow \lambda=-2 \\ & \Rightarrow \overrightarrow{\mathrm{p}}=(-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}) \end{aligned}$ Unit vector $\rightarrow \pm \frac{(-\hat{\mathrm{i}}+\hat{\mathrm{k}})}{\sqrt{2}}$</p>
3. Q3JEE Main 2023 (30 Jan, Shift 2)Scalar Triple Product
   Let $\vec{a}$and $\vec{b}$ be two vectors. Let $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a}\cdot \vec{b}=2$. If $\vec{c}=(2\vec{a}\times \vec{b})-3\vec{b}$, then the value of $\vec{b}\cdot \vec{c}$ is
   1. A.$-24$
   2. B.$-48$
   3. C.$-84$
   4. D.$-60$
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   Answer: (B)

   Given, $\vec{a}$and $\vec{b}$ be two vectors, where $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a}\cdot \vec{b}=2$, Now given, $\vec{c}=\left(2\vec{a}\times \vec{b}\right)-3\vec{b}$ So, $\vec{b}\cdot \vec{c}=\vec{b}\cdot \left(2\vec{a}\times \vec{b}\right)-3\vec{b}\cdot \vec{b}$ $\Rightarrow \vec{b}\cdot \vec{c}=-3|b{|}^{2}$ $\left\{\text{as}\left[\begin{matrix}\vec{b} & 2\vec{a} & \vec{b}\end{matrix}\right]=0\text{or}\vec{b}\cdot \left(2\vec{a}\times \vec{b}\right)=0\right\}$ $\Rightarrow \vec{b}\cdot \vec{c}=-48$
4. Q4JEE Main 2022 (27 Jul, Shift 1)Vector Triple Product
   Let $\vec{a}=2\hat{i}-\hat{j}+5\hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2\hat{k}$. If $\left(\left(\vec{a}\times \vec{b}\right)\times \hat{i}\right)\cdot \hat{k}=\dfrac{23}{2}$, then $\left|\vec{b}\times 2\hat{j}\right|$ is equal to
   1. A.$4$
   2. B.$5$
   3. C.$\sqrt{21}$
   4. D.$\sqrt{17}$
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   Answer: (B)

   Given, $\vec{a}=2\hat{i}-\hat{j}+5\hat{k},\vec{b}=\alpha \hat{i}+\beta \hat{j}+2\hat{k}$ Also given $\left(\left(\vec{a}\times \vec{b}\right)\times \hat{i}\right)\cdot \hat{k}=\dfrac{23}{2}$, then $\left|\vec{b}\times 2\hat{j}\right|$ is Now using triple cross product we have, $\left(\left(\vec{a}\cdot \hat{i}\right)\vec{b}-\left(\vec{b}\cdot \hat{i}\right)\vec{a}\right)\cdot \hat{k}=\dfrac{23}{2}$ $\Rightarrow \left(2.\vec{b}-\alpha \cdot \vec{a}\right).\hat{k}=\dfrac{23}{2}$ $\Rightarrow 2\times 2-\alpha \times 5=\dfrac{23}{2}$ $\Rightarrow 5\alpha =4-\dfrac{23}{2}$ $\Rightarrow \alpha =\dfrac{-3}{2}$ Now finding, $\vec{b}\times 2\hat{j}=\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ \alpha & \beta & 2 \\ 0 & 2 & 0\end{matrix}\right|=-4\hat{i}+2\alpha \hat{k}$ $∴\left|\vec{b}\times 2\hat{j}\right|=\sqrt{16+4{\alpha }^{2}}=\sqrt{16+4\times \dfrac{9}{4}}=5$
5. Q5JEE Main 2026 (05 Apr, Shift 1)Algebra of Vectors
   Let $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ is equal to:
   1. A.$44$
   2. B.$54$
   3. C.$86$
   4. D.$132$
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   Answer: (A)

   Given $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ $\Rightarrow \vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0}$ $\Rightarrow (\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$ $\Rightarrow \vec{r} - \vec{b} = \lambda \vec{a}$ $\Rightarrow \vec{r} = \vec{b} + \lambda \vec{a}$ Taking dot product with $\vec{a}$ on both sides: $\vec{r} \cdot \vec{a} = \vec{b} \cdot \vec{a} + \lambda |\vec{a}|^2$ Since $\vec{r} \cdot \vec{a} = 0$, we get: $\lambda = -\dfrac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}$ We have $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$. $\vec{b} \cdot \vec{a} = (0)(\sqrt{7}) + (1)(1) + (2)(-1) = -1$ $|\vec{a}|^2 = (\sqrt{7})^2 + (1)^2 + (-1)^2 = 9$ $|\vec{b}|^2 = (1)^2 + (2)^2 = 5$ Substituting these values: $\lambda = -\dfrac{-1}{9} = \dfrac{1}{9}$ Therefore, $\vec{r} = \vec{b} + \dfrac{1}{9}\vec{a}$ Squaring both sides: $|\vec{r}|^2 = |\vec{b}|^2 + \dfrac{1}{81}|\vec{a}|^2 + \dfrac{2}{9}(\vec{b} \cdot \vec{a})$ $|\vec{r}|^2 = 5 + \dfrac{1}{81}(9) + \dfrac{2}{9}(-1)$ $|\vec{r}|^2 = 5 + \dfrac{1}{9} - \dfrac{2}{9} = \dfrac{44}{9}$ Finally, $|3\vec{r}|^2 = 9|\vec{r}|^2 = 9 \times \dfrac{44}{9} = 44$ Answer: $44$