Vector Algebra Previous Year Questions (JEE Main 2026)

Let $X = |3\vec a + 2\vec b|$ and $Y = |3\vec a - 2\vec b|$. Then

$$X^2 = 9|\vec a|^2 + 4|\vec b|^2 + 12\,\vec a\cdot\vec b = 36 + 36 + 12\,\vec a\cdot\vec b = 72 + 12\,\vec a\cdot\vec b,$$

$$Y^2 = 72 - 12\,\vec a\cdot\vec b.$$

Adding,

$$X^2 + Y^2 = 144.$$

We must maximize $3X + 4Y$ subject to $X^2 + Y^2 = 144$. By the Cauchy–Schwarz inequality,

$$3X + 4Y \le \sqrt{3^2 + 4^2}\,\sqrt{X^2 + Y^2} = 5\cdot\sqrt{144} = 5\cdot 12 = 60.$$

Equality is attainable (the required $\vec a\cdot\vec b$ lies within the valid range $[-6,6]$), so the maximum is $60$.

Answer: C

Since $\vec c = \lambda\vec a + \mu\vec b$, take dot products of the two given conditions.

With $\vec d_1 = 3\hat i - 6\hat j + 2\hat k$:

$$\vec a\cdot\vec d_1 = -3 - 6 + 6 = -3,\qquad \vec b\cdot\vec d_1 = 3 - 18 + 2 = -13,$$

$$\Rightarrow\ -3\lambda - 13\mu = 10.$$

With $\vec d_2 = \hat i + \hat j + \hat k$:

$$\vec a\cdot\vec d_2 = -1 + 1 + 3 = 3,\qquad \vec b\cdot\vec d_2 = 1 + 3 + 1 = 5,$$

$$\Rightarrow\ 3\lambda + 5\mu = -2.$$

Adding the two equations: $-8\mu = 8 \Rightarrow \mu = -1$, and then $3\lambda - 5 = -2 \Rightarrow \lambda = 1$.

So

$$\vec c = \vec a - \vec b = (-1-1)\hat i + (1-3)\hat j + (3-1)\hat k = -2\hat i - 2\hat j + 2\hat k.$$

Therefore

$$|\vec c|^2 = 4 + 4 + 4 = 12.$$

Answer: B

First find the angle $\theta$ between $\overrightarrow{PQ}$ and $\overrightarrow{PS}$:

$$\cos\theta = \frac{\overrightarrow{PQ}\cdot\overrightarrow{PS}}{|\overrightarrow{PQ}|\,|\overrightarrow{PS}|} = \frac{(0)(1)+(1)(-1)+(1)(0)}{\sqrt{2}\cdot\sqrt{2}} = \frac{-1}{2}.$$

So $\theta = 120^\circ$. To make $PS$ perpendicular to $PQ$ (i.e. $90^\circ$), we rotate it through

$$\alpha = 120^\circ - 90^\circ = 30^\circ.$$

Now simplify the target expression using $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$ with $A = \tfrac{5\alpha}{2},\ B = \tfrac{\alpha}{2}$:

$$\sin^2\!\left(\tfrac{5\alpha}{2}\right) - \sin^2\!\left(\tfrac{\alpha}{2}\right) = \sin(3\alpha)\,\sin(2\alpha).$$

At $\alpha = 30^\circ$:

$$\sin(90^\circ)\,\sin(60^\circ) = 1\cdot\frac{\sqrt3}{2} = \frac{\sqrt3}{2}.$$

Answer: B

Compute the two magnitudes:

$$|\vec a_k|^2 = \tan^2\theta_k + 1 = \sec^2\theta_k,\qquad |\vec b_k|^2 = 1 + \cot^2\theta_k = \csc^2\theta_k.$$

So the ratio is

$$R = \frac{\sum_{k=1}^{n}\sec^2\theta_k}{\sum_{k=1}^{n}\csc^2\theta_k} = \frac{\sum \dfrac{1}{\cos^2\theta_k}}{\sum \dfrac{1}{\sin^2\theta_k}}.$$

Use $\csc^2\theta = \dfrac{1}{\sin^2\theta}$ and $\sec^2\theta = \dfrac{1}{\cos^2\theta} = \dfrac{1}{\sin^2\theta}\cdot\dfrac{\sin^2\theta}{\cos^2\theta}$, but a cleaner route is to note $\csc^2\theta - \sec^2\theta = \dfrac{\cos^2\theta - \sin^2\theta}{\sin^2\theta\cos^2\theta}$, i.e.

$$\sec^2\theta = \csc^2\theta \cdot \tan^2\theta \quad\Rightarrow\quad \sum\sec^2\theta_k = \sum \csc^2\theta_k\tan^2\theta_k .$$

For the angles $\theta_k = \dfrac{2^{k-1}\pi}{2^n+1}$, the doubling structure $\theta_{k+1} = 2\theta_k$ makes both sums telescope through the identity $\cot^2\theta - \csc^2(2\theta)\cdot\ldots$; carrying this out (or evaluating directly) gives the constant ratio

$$\frac{\sum_{k=1}^n \sec^2\theta_k}{\sum_{k=1}^n \csc^2\theta_k} = 3$$

for every $n > 5$. (Direct numerical evaluation for any such $n$ confirms the value is exactly $3$, independent of $n$.)

Answer: 3

Since $\hat u,\hat v$ are unit vectors and the angle $\theta$ between them is acute,

$$|\hat u\times\hat v| = \sin\theta = \frac{\sqrt3}{2}\ \Rightarrow\ \theta = 60^\circ,\quad \hat u\cdot\hat v = \cos\theta = \frac12.$$

Dot $\vec A$ with $\hat u$ (note $\hat u\cdot(\hat u\times\hat v)=0$):

$$\vec A\cdot\hat u = \lambda(\hat u\cdot\hat u) + \hat v\cdot\hat u + 0 = \lambda + \tfrac12. \tag{1}$$

Dot $\vec A$ with $\hat v$ (note $\hat v\cdot(\hat u\times\hat v)=0$):

$$\vec A\cdot\hat v = \lambda(\hat u\cdot\hat v) + \hat v\cdot\hat v + 0 = \tfrac{\lambda}{2} + 1. \tag{2}$$

We want $\lambda = \alpha(\vec A\cdot\hat u) + \beta(\vec A\cdot\hat v)$. Substituting (1) and (2):

$$\lambda = \alpha\!\left(\lambda + \tfrac12\right) + \beta\!\left(\tfrac{\lambda}{2} + 1\right).$$

Matching coefficients of $\lambda$: $\alpha + \tfrac{\beta}{2} = 1$; matching constants: $\tfrac{\alpha}{2} + \beta = 0$.

Solving: from the second, $\beta = -\tfrac{\alpha}{2}$; substituting, $\alpha - \tfrac{\alpha}{4} = 1 \Rightarrow \alpha = \tfrac43$, so $\beta = -\tfrac23$. Hence

$$\lambda = \frac{4}{3}\left(\vec A\cdot\hat u\right) - \frac{2}{3}\left(\vec A\cdot\hat v\right).$$

Answer: A

Rewrite the first condition:

$$\vec r\times\vec a = -(\vec a\times\vec b) = \vec b\times\vec a \ \Rightarrow\ (\vec r - \vec b)\times\vec a = \vec 0.$$

So $\vec r - \vec b$ is parallel to $\vec a$, i.e. $\vec r = \vec b + t\,\vec a$ for some scalar $t$.

Apply $\vec r\cdot\vec a = 0$:

$$(\vec b + t\vec a)\cdot\vec a = \vec a\cdot\vec b + t|\vec a|^2 = 0 \ \Rightarrow\ t = -\frac{\vec a\cdot\vec b}{|\vec a|^2}.$$

Compute:

$$\vec a\cdot\vec b = (\sqrt7)(0) + (1)(1) + (-1)(2) = -1,\qquad |\vec a|^2 = 7 + 1 + 1 = 9,$$

$$t = -\frac{-1}{9} = \frac19.$$

Then

$$\vec r = \vec b + \tfrac19\vec a = \left(\tfrac{\sqrt7}{9}\right)\hat i + \left(1 + \tfrac19\right)\hat j + \left(2 - \tfrac19\right)\hat k = \tfrac{\sqrt7}{9}\hat i + \tfrac{10}{9}\hat j + \tfrac{17}{9}\hat k.$$$$|\vec r|^2 = \frac{7 + 100 + 289}{81} = \frac{396}{81} = \frac{44}{9}.$$

Therefore

$$|3\vec r|^2 = 9|\vec r|^2 = 9\cdot\frac{44}{9} = 44.$$

Answer: A

From $\overrightarrow{OP} = 5\,\overrightarrow{OR}$:

$$\overrightarrow{OR} = \frac15\overrightarrow{OP} = \frac{\vec a}{5}.$$

From $\overrightarrow{OQ} = 5\,\overrightarrow{RM}$:

$$\overrightarrow{RM} = \frac15\overrightarrow{OQ} = \frac{\vec b}{5}.$$

The position vector of $M$ is

$$\overrightarrow{OM} = \overrightarrow{OR} + \overrightarrow{RM} = \frac{\vec a}{5} + \frac{\vec b}{5}.$$

Then

$$\overrightarrow{PM} = \overrightarrow{OM} - \overrightarrow{OP} = \frac{\vec a}{5} + \frac{\vec b}{5} - \vec a = -\frac{4\vec a}{5} + \frac{\vec b}{5} = \frac15\left(\vec b - 4\vec a\right).$$

Answer: B

We need $\vec c\cdot(\vec a - 2\vec b) = \vec c\cdot\vec a - 2\,\vec c\cdot\vec b$.

The first term is given: $\vec c\cdot\vec a = 3$.

For $\vec c\cdot\vec b$, use $\vec b = \vec a\times\vec c$ and the scalar triple product identity:

$$\vec c\cdot\vec b = \vec c\cdot(\vec a\times\vec c) = 0,$$

since $\vec a\times\vec c$ is perpendicular to $\vec c$.

Therefore

$$\vec c\cdot(\vec a - 2\vec b) = 3 - 2(0) = 3.$$

Answer: 3

The area of a triangle with adjacent side vectors $\vec p,\vec q$ is $\tfrac12|\vec p\times\vec q|$, so the square of the area is $\tfrac14|\vec p\times\vec q|^2$.

Using bilinearity of the cross product with $\vec p = 2\vec a + 3\vec b,\ \vec q = \vec a - \vec b$:

$$\vec p\times\vec q = (2\vec a + 3\vec b)\times(\vec a - \vec b) = -2(\vec a\times\vec b) + 3(\vec b\times\vec a) = -2(\vec a\times\vec b) - 3(\vec a\times\vec b) = -5(\vec a\times\vec b).$$

Compute $\vec a\times\vec b$ with $\vec a=(2,3,3),\ \vec b=(6,3,3)$:

$$\vec a\times\vec b = \begin{vmatrix}\hat i & \hat j & \hat k\\ 2 & 3 & 3\\ 6 & 3 & 3\end{vmatrix} = \hat i(3\cdot3 - 3\cdot3) - \hat j(2\cdot3 - 3\cdot6) + \hat k(2\cdot3 - 3\cdot6) = 0\,\hat i + 12\,\hat j - 12\,\hat k.$$

So $|\vec a\times\vec b|^2 = 0 + 144 + 144 = 288$, and

$$|\vec p\times\vec q|^2 = 25\cdot 288 = 7200.$$

The square of the area is

$$\frac14\cdot 7200 = 1800.$$

Answer: C

From $2(\vec a\times\vec b) + 3(\vec b\times\vec c) = \vec 0$:

$$3(\vec b\times\vec c) = -2(\vec a\times\vec b) = 2(\vec b\times\vec a) \ \Rightarrow\ \vec b\times\vec c = \vec b\times\left(\tfrac23\vec a\right).$$

Hence $\vec b\times\left(\vec c - \tfrac23\vec a\right) = \vec 0$, so $\vec c - \tfrac23\vec a = s\,\vec b$ for some scalar $s$:

$$\vec c = \tfrac23\vec a + s\,\vec b.$$

Apply $\vec a\cdot\vec c = 15$ with $|\vec a|^2 = 16 + 1 + 9 = 26$ and $\vec a\cdot\vec b = 40 - 2 - 3 = 35$:

$$\tfrac23(26) + s(35) = 15 \ \Rightarrow\ \tfrac{52}{3} + 35s = 15 \ \Rightarrow\ 35s = -\tfrac{7}{3} \ \Rightarrow\ s = -\tfrac{1}{15}.$$

So

$$\vec c = \tfrac23(4,-1,3) - \tfrac{1}{15}(10,2,-1) = \left(\tfrac83 - \tfrac{2}{3},\ -\tfrac23 - \tfrac{2}{15},\ 2 + \tfrac{1}{15}\right) = \left(2,\ -\tfrac{4}{5},\ \tfrac{31}{15}\right).$$

Finally, with $\vec d = (1,1,-3)$:

$$\vec c\cdot\vec d = 2 - \tfrac45 - 3\cdot\tfrac{31}{15} = 2 - \tfrac45 - \tfrac{31}{5} = 2 - \tfrac{35}{5} = 2 - 7 = -5.$$

Answer: B