JEE PYQ: Vector Algebra - Question ID 0fe036d3e070 (JEE Main 2004)

In vector algebra, the cross product (or vector product) of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is defined as: $\overrightarrow{A} \times \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \sin \theta \ \hat{n}$ where: - $|\overrightarrow{A}|$ and $|\overrightarrow{B}|$ are the magnitudes of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, - $\theta$ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$, - $\hat{n}$ is a unit vector perpendicular to the plane containing $\overrightarrow{A}$ and $\overrightarrow{B}$, following the right-hand rule.

A key property of the cross product is its anti-commutative nature: $\overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{B} \times \overrightarrow{A})$ This means that swapping the order of the vectors reverses the direction of the resulting vector.

The question states that $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$. Using the anti-commutative property, this implies: $\overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{A} \times \overrightarrow{B})$ Let $\overrightarrow{C} = \overrightarrow{A} \times \overrightarrow{B}$. Then the equation becomes: $\overrightarrow{C} = -\overrightarrow{C}$ Adding $\overrightarrow{C}$ to both sides gives: $2\overrightarrow{C} = \overrightarrow{0} \implies \overrightarrow{C} = \overrightarrow{0}$ Thus, $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{0}$.

The cross product $\overrightarrow{A} \times \overrightarrow{B}$ is zero if and only if: 1. Either $\overrightarrow{A}$ or $\overrightarrow{B}$ is the zero vector, or 2. The angle $\theta$ between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $0$ or $\pi$ (i.e., the vectors are parallel or anti-parallel).

Since the question does not specify that either vector is zero, the only remaining possibility is that the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\pi$ (or $0$, but $\pi$ is the more general case of anti-parallel vectors).

1. Start with the given condition: $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$
2. Apply the anti-commutative property of the cross product: $\overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{A} \times \overrightarrow{B})$
3. Let $\overrightarrow{C} = \overrightarrow{A} \times \overrightarrow{B}$. Substitute into the equation: $\overrightarrow{C} = -\overrightarrow{C}$
4. Add $\overrightarrow{C}$ to both sides: $2\overrightarrow{C} = \overrightarrow{0}$
5. Divide by 2: $\overrightarrow{C} = \overrightarrow{0}$ Thus: $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{0}$
6. Recall the definition of the cross product: $\overrightarrow{A} \times \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \sin \theta \ \hat{n} = \overrightarrow{0}$ For this to hold, either:
   • $|\overrightarrow{A}| = 0$ or $|\overrightarrow{B}| = 0$ (trivial case, not considered here), or
   • $\sin \theta = 0$.
7. Solve $\sin \theta = 0$: $\theta = 0, \pi, 2\pi, \dots$ Since angles are typically considered in the range $[0, \pi]$, the possible solutions are $\theta = 0$ or $\theta = \pi$.
8. $\theta = 0$ corresponds to parallel vectors, while $\theta = \pi$ corresponds to anti-parallel vectors. Both cases satisfy $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{0}$, but the question asks for the angle between the vectors, and $\pi$ is the more general answer (as it includes the case where the vectors are in opposite directions).
9. Among the given options:
   • A: $\pi/2$ (90°) → Incorrect, as $\sin(\pi/2) = 1 \neq 0$.
   • B: $\pi/3$ (60°) → Incorrect, as $\sin(\pi/3) \neq 0$.
   • C: $\pi$ (180°) → Correct, as $\sin(\pi) = 0$.
   • D: $\pi/4$ (45°) → Incorrect, as $\sin(\pi/4) \neq 0$.

Students often make the following mistakes in this question:

1. Ignoring the anti-commutative property: Some students forget that $\overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{B} \times \overrightarrow{A})$ and incorrectly assume that the cross product is commutative. This leads them to think that the given condition is always true, which is not the case.
2. Overlooking the zero vector case: While the question does not specify that the vectors are non-zero, the options provided suggest that the angle is the key factor. Students should recognize that the cross product is zero only when the vectors are parallel or anti-parallel (or one of them is zero).
3. Confusing $\theta = 0$ and $\theta = \pi$: Both angles satisfy $\sin \theta = 0$, but the question asks for the angle between the vectors. While $\theta = 0$ is a valid solution, $\theta = \pi$ is the more general answer (as it includes the case where the vectors are in opposite directions). The options only include $\pi$, so this is the correct choice.
4. Misapplying the right-hand rule: Some students might think that the cross product being equal implies that the vectors are perpendicular (i.e., $\theta = \pi/2$). However, this is incorrect because the cross product is zero (not equal) when the vectors are perpendicular to their own plane (which is not the case here).

Exam Tip: Always recall the fundamental properties of vector operations. For the cross product, remember:

• It is anti-commutative: $\overrightarrow{A} \times \overrightarrow{B} = - (\overrightarrow{B} \times \overrightarrow{A})$.
• It is zero if and only if the vectors are parallel or anti-parallel (or one of them is zero).
• The magnitude of the cross product is $|\overrightarrow{A}| |\overrightarrow{B}| \sin \theta$.