JEE PYQ: Vector Algebra - Question ID a6ccb73b1805 (JEE Main 2021)

Core Formula & Concept:

In vector algebra, the cross product (or vector product) of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is defined as: $\overrightarrow{P} \times \overrightarrow{Q} = |\overrightarrow{P}| |\overrightarrow{Q}| \sin \theta \ \hat{n}$ where:

• $\theta$ is the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$ ($0^\circ \leq \theta \leq 180^\circ$),
• $\hat{n}$ is a unit vector perpendicular to the plane containing $\overrightarrow{P}$ and $\overrightarrow{Q}$, following the right-hand rule.

A fundamental property of the cross product is its anti-commutativity: $\overrightarrow{P} \times \overrightarrow{Q} = - (\overrightarrow{Q} \times \overrightarrow{P})$ This means the cross product changes sign when the order of the vectors is reversed.

Step-by-Step Derivation:

Given the equation: $\overrightarrow{P} \times \overrightarrow{Q} = \overrightarrow{Q} \times \overrightarrow{P}$ Using the anti-commutative property of the cross product, we substitute: $\overrightarrow{P} \times \overrightarrow{Q} = - (\overrightarrow{P} \times \overrightarrow{Q})$ Let $\overrightarrow{R} = \overrightarrow{P} \times \overrightarrow{Q}$. Then the equation becomes: $\overrightarrow{R} = -\overrightarrow{R}$ Add $\overrightarrow{R}$ to both sides: $\overrightarrow{R} + \overrightarrow{R} = \overrightarrow{0} \implies 2\overrightarrow{R} = \overrightarrow{0}$ Thus: $\overrightarrow{R} = \overrightarrow{0}$ Substituting back the definition of $\overrightarrow{R}$: $\overrightarrow{P} \times \overrightarrow{Q} = \overrightarrow{0}$ From the definition of the cross product, this implies: $|\overrightarrow{P}| |\overrightarrow{Q}| \sin \theta = 0$ Since the problem specifies $0^\circ < \theta < 360^\circ$, and assuming $\overrightarrow{P}$ and $\overrightarrow{Q}$ are non-zero vectors (otherwise the angle is undefined), we have: $\sin \theta = 0$ The solutions to $\sin \theta = 0$ in the interval $0^\circ < \theta < 360^\circ$ are: $\theta = 180^\circ$ (Note: $\theta = 0^\circ$ is excluded by the problem's condition $0^\circ < \theta < 360^\circ$, and $\theta = 360^\circ$ is equivalent to $0^\circ$.) Thus, the only valid solution is $\theta = 180^\circ$.

Common Traps & Exam Tip:

1. Ignoring the anti-commutative property: Many students forget that $\overrightarrow{P} \times \overrightarrow{Q} \neq \overrightarrow{Q} \times \overrightarrow{P}$ and incorrectly assume the cross product is commutative. This leads to missing the key step where the equation simplifies to $\overrightarrow{P} \times \overrightarrow{Q} = \overrightarrow{0}$.
2. Overlooking the zero vector condition: Students may stop at $\sin \theta = 0$ and consider $\theta = 0^\circ$ or $180^\circ$ without checking the problem's constraints ($0^\circ < \theta < 360^\circ$). The problem explicitly excludes $\theta = 0^\circ$, so only $\theta = 180^\circ$ is valid.
3. Assuming parallel vectors: Some students conclude that $\overrightarrow{P}$ and $\overrightarrow{Q}$ are parallel (which is true for $\theta = 0^\circ$ or $180^\circ$) but fail to eliminate $\theta = 0^\circ$ due to the problem's condition.
4. Sign errors in cross product: Misapplying the right-hand rule or the direction of $\hat{n}$ can lead to confusion, but this is not directly relevant here since the magnitudes cancel out.

Exam Tip: Always recall the anti-commutative nature of the cross product. If an equation like $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$ appears, immediately recognize that it implies $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{0}$.