Vector Algebra: JEE Main 2026 PYQs

Answer: Option (3)

Key idea: Vector Magnitude Optimization using Triangle Inequality and Calculus

1. Express $|3\vec{a} + 2\vec{b}|^2 = 72(1 + \cos\theta)$ and $|3\vec{a} - 2\vec{b}|^2 = 72(1 - \cos\theta)$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$
2. Substitute to get $f(\theta) = 18\sqrt{2(1 + \cos\theta)} + 24\sqrt{2(1 - \cos\theta)}$
3. Let $x = \cos\theta$ and find critical points: $f'(x) = \frac{9\sqrt{2}}{\sqrt{1+x}} - \frac{12\sqrt{2}}{\sqrt{1-x}} = 0$
4. Solve to get $9\sqrt{1-x} = 12\sqrt{1+x}$, which gives $\cos\theta = -\frac{7}{25}$
5. Calculate $f(-\frac{7}{25}) = 18 \cdot \frac{6}{5} + 24 \cdot \frac{8}{5} = \frac{108 + 192}{5} = 60$
6. Verify this exceeds boundary values at $\theta = 0$ and $\theta = \pi$

Answer: Option (2)

Key idea: Vector operations, dot product, and solving linear equations to find vector magnitude

1. Express $\vec{c} = \lambda \vec{a} + \mu \vec{b} = (-\lambda + \mu)\hat{i} + (\lambda + 3\mu)\hat{j} + (3\lambda + \mu)\hat{k}$
2. Use condition $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ to get $-3\lambda - 13\mu = 10$
3. Use condition $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$ to get $3\lambda + 5\mu = -2$
4. Solve the system: Add equations to get $-8\mu = 8$, so $\mu = -1$ and $\lambda = 1$
5. Find $\vec{c} = \vec{a} - \vec{b} = -2\hat{i} - 2\hat{j} + 2\hat{k}$
6. Calculate $|\vec{c}|^2 = (-2)^2 + (-2)^2 + (2)^2 = 12$

Answer: Option (2)

Key idea: Vector dot product, angle between vectors, rotation in plane, trigonometric identities

1. Find the angle between $\overrightarrow{PQ}$ and $\overrightarrow{PS}$ using $\cos\theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PS}}{|\overrightarrow{PQ}||\overrightarrow{PS}|} = \frac{-1}{2}$, so $\theta = 120°$
2. After rotation by $\alpha$, vectors become perpendicular (angle = $90°$), so $120° - \alpha = 90°$, giving $\alpha = 30°$
3. Calculate $\frac{\alpha}{2} = 15°$ and $\frac{5\alpha}{2} = 75°$
4. Use identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$
5. Evaluate: $\sin^2(75°) - \sin^2(15°) = \sin(90°)\sin(60°) = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

Answer: 1

Key idea: Vector magnitude, trigonometric identities, and summation of trigonometric series

1. Calculate $|\vec{a_k}|^2 = \tan^2 \theta_k + 1 = \sec^2 \theta_k$
2. Calculate $|\vec{b_k}|^2 = 1 + \cot^2 \theta_k = \csc^2 \theta_k$
3. The ratio becomes $\frac{\sum_{k=1}^{n} \sec^2 \theta_k}{\sum_{k=1}^{n} \csc^2 \theta_k}$
4. Note that $\theta_k = \frac{(2^k - 1)\pi}{2^{n+1}}$ gives angles with special symmetry properties
5. Using the symmetry of the angles around $\pi/4$ and properties of the specific sequence, the sums are equal
6. Therefore, the ratio equals $1$

Answer: Option (1)

Key idea: Vector dot and cross products with unit vectors

1. Find the angle: Since $|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$, we have $\sin\theta = \frac{\sqrt{3}}{2}$, so $\theta = 60°$ and $\hat{u} \cdot \hat{v} = \cos 60° = \frac{1}{2}$
2. Compute $\vec{A} \cdot \hat{u}$: Using $\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})$ and noting $(\hat{u} \times \hat{v}) \perp \hat{u}$, we get $\vec{A} \cdot \hat{u} = \lambda + \frac{1}{2}$
3. Compute $\vec{A} \cdot \hat{v}$: Similarly, $(\hat{u} \times \hat{v}) \perp \hat{v}$, so $\vec{A} \cdot \hat{v} = \frac{\lambda}{2} + 1$
4. Test option (1): Calculate $\frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v}) = \frac{4}{3}(\lambda + \frac{1}{2}) - \frac{2}{3}(\frac{\lambda}{2} + 1)$
5. Simplify: $= \frac{4\lambda}{3} + \frac{2}{3} - \frac{\lambda}{3} - \frac{2}{3} = \frac{3\lambda}{3} = \lambda$ ✓

Answer: Option (1)

Key idea: Vector cross product and dot product properties

1. Given $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$, rewrite as $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$
2. This implies $(\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$, so $\vec{r} = \vec{b} + \lambda \vec{a}$ for some scalar $\lambda$
3. Use condition $\vec{r} \cdot \vec{a} = 0$: $(\vec{b} + \lambda \vec{a}) \cdot \vec{a} = 0$ gives $\vec{b} \cdot \vec{a} + \lambda |\vec{a}|^2 = 0$
4. Calculate $\vec{b} \cdot \vec{a} = -1$ and $|\vec{a}|^2 = 9$, so $\lambda = \frac{1}{9}$
5. Find $\vec{r} = \vec{b} + \frac{1}{9}\vec{a} = \frac{\sqrt{7}}{9}\hat{i} + \frac{10}{9}\hat{j} + \frac{17}{9}\hat{k}$
6. Calculate $|\vec{r}|^2 = \frac{7 + 100 + 289}{81} = \frac{396}{81} = \frac{44}{9}$
7. Therefore $|3\vec{r}|^2 = 9|\vec{r}|^2 = 9 \times \frac{44}{9} = 44$

Answer: Option (1)

Key idea: Position vectors and vector addition using triangle law

1. Find $\overrightarrow{OR}$: Since $\overrightarrow{OP} = 5\overrightarrow{OR}$, we get $\overrightarrow{OR} = \frac{\vec{a}}{5}$
2. Find $\overrightarrow{RM}$: Since $\overrightarrow{OQ} = 5\overrightarrow{RM}$, we get $\overrightarrow{RM} = \frac{\vec{b}}{5}$
3. Find $\overrightarrow{OM}$: Using $\overrightarrow{OM} = \overrightarrow{OR} + \overrightarrow{RM}$, we get $\overrightarrow{OM} = \frac{\vec{a} + \vec{b}}{5}$
4. Find $\overrightarrow{PM}$: Using $\overrightarrow{PM} = \overrightarrow{OM} - \overrightarrow{OP}$
5. Substitute: $\overrightarrow{PM} = \frac{\vec{a} + \vec{b}}{5} - \vec{a} = \frac{\vec{b} - 4\vec{a}}{5}$

Answer: 3

Key idea: Vector cross product, dot product, and BAC-CAB rule

1. Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$, $\vec{a} \times \vec{c} = \vec{b}$, and $\vec{a} \cdot \vec{c} = 3$
2. Apply BAC-CAB rule: $\vec{a} \times (\vec{a} \times \vec{c}) = \vec{a}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{a})$, which gives $\vec{a} \times \vec{b} = 3\vec{a} - 3\vec{c}$
3. Calculate $\vec{a} \times \vec{b} = -2\hat{i} + \hat{j} + \hat{k}$ using determinant
4. Solve for $\vec{c}$: $3\vec{c} = 3\vec{a} - (\vec{a} \times \vec{b}) = 5\hat{i} + 2\hat{j} + 2\hat{k}$, so $\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$
5. Calculate $\vec{c} \cdot \vec{b} = \frac{5}{3}(0) + \frac{2}{3}(1) + \frac{2}{3}(-1) = 0$
6. Find $\vec{c} \cdot (\vec{a} - 2\vec{b}) = \vec{c} \cdot \vec{a} - 2\vec{c} \cdot \vec{b} = 3 - 2(0) = 3$

Answer: 1800

Key idea: Area of triangle using cross product of vectors

1. Step 1: Calculate $2\vec{a} + 3\vec{b} = 4\hat{i} + 6\hat{j} + 6\hat{k} + 18\hat{i} + 9\hat{j} + 9\hat{k} = 22\hat{i} + 15\hat{j} + 15\hat{k}$
2. Step 2: Calculate $\vec{a} - \vec{b} = (2-6)\hat{i} + (3-3)\hat{j} + (3-3)\hat{k} = -4\hat{i}$
3. Step 3: Find cross product $(2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 22 & 15 & 15 \\ -4 & 0 & 0 \end{vmatrix} = -60\hat{j} + 60\hat{k}$
4. Step 4: Calculate magnitude $|\vec{u} \times \vec{v}| = \sqrt{60^2 + 60^2} = \sqrt{7200} = 60\sqrt{2}$
5. Step 5: Area of triangle $= \frac{1}{2}|\vec{u} \times \vec{v}| = \frac{1}{2} \cdot 60\sqrt{2} = 30\sqrt{2}$
6. Step 6: Square of area $= (30\sqrt{2})^2 = 900 \times 2 = 1800$

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