Vector Algebra JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Vector Algebra, free to read — no sign-in needed. The full chapter has 46 questions; sign in to attempt the remaining 41 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Let $\vec{a}, \vec{b}$ be two vectors, and let $P, Q$ and $R$ be the points with position vectors $\vec{a}$, $\vec{b}$ and $\vec{a} + \vec{b}$, respectively, with respect to the origin $O$. If $|\vec{a} + \vec{b}| = \sqrt{21}$, $|\vec{a} - \vec{b}| = 3$, and $\vec{a}$ and $(\vec{a} - \vec{b})$ are perpendicular to each other, then the area of the triangle $OPR$ is
   1. A.$\sqrt{3}$
   2. B.$\dfrac{\sqrt{3}}{2}$
   3. C.$\dfrac{3\sqrt{3}}{2}$
   4. D.$\dfrac{3}{2}$
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   Answer: (C)

   Given $|\vec{a} + \vec{b}| = \sqrt{21}$ and $|\vec{a} - \vec{b}| = 3$. Squaring both equations: $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 21$ $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 9$ Subtracting the second equation from the first gives: $4\vec{a} \cdot \vec{b} = 12 \Rightarrow \vec{a} \cdot \vec{b} = 3$ Adding the two equations gives: $2(|\vec{a}|^2 + |\vec{b}|^2) = 30 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 = 15$ Since $\vec{a}$ and $(\vec{a} - \vec{b})$ are perpendicular, their dot product is zero: $\vec{a} \cdot (\vec{a} - \vec{b}) = 0 \Rightarrow |\vec{a}|^2 - \vec{a} \cdot \vec{b} = 0 \Rightarrow |\vec{a}|^2 = \vec{a} \cdot \vec{b} = 3$ Substituting $|\vec{a}|^2 = 3$ into $|\vec{a}|^2 + |\vec{b}|^2 = 15$ gives: $3 + |\vec{b}|^2 = 15 \Rightarrow |\vec{b}|^2 = 12$ Using Lagrange's identity to find $|\vec{a} \times \vec{b}|$: $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$ $|\vec{a} \times \vec{b}|^2 = (3)(12) - (3)^2 = 36 - 9 = 27$ $|\vec{a} \times \vec{b}| = \sqrt{27} = 3\sqrt{3}$ The position vectors of $P$ and $R$ are $\vec{a}$ and $\vec{a} + \vec{b}$ respectively. The area of triangle $OPR$ is: $\dfrac{1}{2} |\vec{OP} \times \vec{OR}| = \dfrac{1}{2} |\vec{a} \times (\vec{a} + \vec{b})| = \dfrac{1}{2} |\vec{a} \times \vec{a} + \vec{a} \times \vec{b}| = \dfrac{1}{2} |\vec{a} \times \vec{b}|$ Substituting the value of $|\vec{a} \times \vec{b}|$: Area = $\dfrac{3\sqrt{3}}{2}$ Answer: $\dfrac{3\sqrt{3}}{2}$
2. Q2JEE Advanced Adv 2025 (Paper 2)
   Consider the vectors $\vec{x}=\hat{i}+2 \hat{j}+3 \hat{k}, \quad \vec{y}=2 \hat{i}+3 \hat{j}+\hat{k}, \quad \text { and } \quad \vec{z}=3 \hat{i}+\hat{j}+2 \hat{k}$ For two distinct positive real numbers $\alpha$ and $\beta$, define $\vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z}, \quad \vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x}, \quad \text { and } \quad \vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y}$ If the vectors $\vec{X}, \vec{Y}$, and $\vec{Z}$ lie in a plane, the value of $\alpha+\beta-3$ is ______ .
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   Answer: -2

   $\Rightarrow\left|\begin{array}{ccc}\alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha\end{array}\right| \underbrace{\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{array}\right|}_{-0}=0$ $\begin{aligned} & \Rightarrow\left(\alpha^3+\beta^3-1\right)-(-\alpha \beta-\alpha \beta-\alpha \beta)=0 \\ & \Rightarrow \alpha^3+\beta^3+3 \alpha \beta=1 \\ & \Rightarrow \alpha^3+\beta^3+(-1)^3=3(\alpha)(\beta)(-1) \\ & \Rightarrow \alpha+\beta-1=0\end{aligned}$ So, $\alpha+\beta-3=-2$
3. Q3JEE Advanced Adv 2024 (Paper 2)
   Let $\vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$. If for some real numbers $\alpha, \beta$, and $\gamma$, we have $15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} \times \vec{q})$ then the value of $\gamma$ is
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   Answer: 2

   $15 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}=\alpha(2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})+\beta(\overrightarrow{\mathrm{p}}-2 \overrightarrow{\mathrm{q}})+\gamma(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})$ taking dot with $(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})$ $\left|\begin{array}{ccc}15 & 10 & 6 \\ 2 & 1 & 3 \\ 1 & -1 & 1\end{array}\right|=0+0+\gamma\left(\mathrm{p}^2 \mathrm{q}^2-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2\right) \quad\left[\because(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})^2+(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2=\mathrm{p}^2 \mathrm{q}^2\right]$ $\begin{aligned} & \Rightarrow 52=26 \gamma \\ & \therefore \gamma=2\end{aligned}$
4. Q4JEE Advanced Adv 2024 (Paper 1)
   Let $\overrightarrow{O P}=\frac{\alpha-1}{\alpha} \hat{i}+\hat{j}+\hat{k}, \overrightarrow{O Q}=\hat{i}+\frac{\beta-1}{\beta} \hat{j}+\hat{k}$ and $\overrightarrow{O R}=\hat{i}+\hat{j}+\frac{1}{2} \hat{k}$ be three vectors, where $\alpha, \beta \in \mathbb{R}-\{0\}$ and $O$ denotes the origin. If $(\overrightarrow{O P} \times \overrightarrow{O Q}) \cdot \overrightarrow{O R}=0$ and the point $(\alpha, \beta, 2)$ lies on the plane $3 x+3 y-z+l=0$, then the value of $l$ is ____
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   Answer: 5

   $(\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OQ}}) \cdot \overrightarrow{\mathrm{OR}}=0$ $\left|\begin{array}{ccc}\frac{\alpha-1}{\alpha} & 1 & 1 \\ 1 & \frac{\beta-1}{\beta} & 1 \\ 1 & 1 & \frac{1}{2}\end{array}\right|=0$ $\alpha+\beta+1=0...(1)$ Also $\quad(\alpha, \beta, 2)$ lies on $3 \mathrm{x}+3 \mathrm{y}-\mathrm{z}+l=0$ $\Rightarrow \quad 3 \alpha+3 \beta-2+l=0 \quad \Rightarrow \quad l=2-3(\alpha+\beta)$ ...(2) use (1) in (2) $\Rightarrow l=5$
5. Q5JEE Advanced Adv 2023 (Paper 2)
   Let the position vectors of points $P,Q,R$ and $S$ be $\vec{a}=\hat{i}+2\hat{j}-5\hat{k},\vec{b}=3\hat{i}+6\hat{j}+3\hat{k},\vec{c}=\dfrac{17}{5}\hat{i}+\dfrac{16}{5}\hat{j}+7\hat{k}$ and $\vec{d}=2\hat{i}+\hat{j}+\hat{k}$, respectively. Then which of the following statements is true?
   1. A.The points $P,Q,R$ and $S$ are NOT coplanar
   2. B.$\dfrac{\vec{b}+2\vec{d}}{3}$ is the position vector of a point which divides $PR$ internally in the ratio $5:4$
   3. C.$\dfrac{\vec{b}+2\vec{d}}{3}$ is the position vector of a point which divides $PR$ externally in the ratio $5:4$
   4. D.The square of magnitude of the vector $\vec{b}\times \vec{d}$ is $95$
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   Answer: (B)

   Given, $P(1,2,-5),Q(3,6,3),R\left(\dfrac{17}{5},\dfrac{16}{5},7\right)\&S(2,1,1)$ Now solving options, Option $\left(A\right)$ checking points are coplanar, $\vec{PQ}=\left(2,4,8\right),\vec{PR}=\left(\dfrac{12}{5},\dfrac{6}{5},12\right)\&\vec{PQ}=\left(1,-1,6\right)$ Now if points are coplanar then, $\left[\begin{matrix}\vec{PQ} & \vec{PR} & \vec{PS}\end{matrix}\right]=0$ Now solving L.H.S, we get, $\left|\begin{matrix}2 & 4 & 8 \\ \dfrac{12}{5} & \dfrac{6}{5} & 12 \\ 1 & -1 & 6\end{matrix}\right|=\dfrac{2}{5}\left|\begin{matrix}1 & 2 & 4 \\ 12 & 6 & 60 \\ 1 & -1 & 6\end{matrix}\right|=\dfrac{2}{5}\left(96-24-72\right)=0$ Hence, they are coplanar So option $\left(A\right)$ is wrong. Now solving, option $\left(B\right)$ we get, $\dfrac{\vec{b}+2\vec{d}}{3}=\dfrac{7i+8j+5k}{3}$ Now let point divides the $PR\text{in}\lambda :1$ we get, So, on taking coefficient of $\hat{i}$ we get, $\Rightarrow \dfrac{\dfrac{17\lambda }{5}+1}{(\lambda +1)}=\dfrac{7}{3}$ $\Rightarrow \dfrac{17\lambda }{5}+1=\dfrac{7}{3}(\lambda +1)$ $\Rightarrow 51\lambda +15=35\lambda +35$ $\Rightarrow 16\lambda =20$ $\Rightarrow \lambda =\dfrac{5}{4}$, hence option $\left(B\right)$ is correct and option $\left(C\right)$ is wrong, Now solving option $\left(D\right)$ we get, $\vec{b}\times \vec{d}=\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 6 & 3 \\ 2 & 1 & 1\end{matrix}\right|=3\hat{i}+3\hat{j}-9\hat{k}$ $\Rightarrow {\left|\vec{b}\times \vec{d}\right|}^{2}={\left(\sqrt{{3}^{2}+{3}^{2}+{9}^{2}}\right)}^{2}=99$, Hence, option $\left(D\right)$ is wrong.