JEE PYQ: Vector Algebra - Question ID 5c7bcce0f284 (JEE Main 2022)

In vector algebra, two fundamental operations on a vector $\overrightarrow{A}$ are:

• Dot Product (Scalar Product): The dot product of a vector with itself is defined as $\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}|^2 \cos \theta$ where $\theta$ is the angle between the two vectors. Since $\overrightarrow{A}$ is being dotted with itself, $\theta = 0^\circ$ and $\cos 0^\circ = 1$. Thus, $\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}|^2$ This is always non-negative and equals zero only if $|\overrightarrow{A}| = 0$.


• Cross Product (Vector Product): The cross product of a vector with itself is defined as $\overrightarrow{A} \times \overrightarrow{A} = |\overrightarrow{A}| |\overrightarrow{A}| \sin \theta \hat{n}$ where $\hat{n}$ is the unit vector perpendicular to the plane containing $\overrightarrow{A}$ and $\overrightarrow{A}$. Again, since the angle $\theta$ between $\overrightarrow{A}$ and itself is $0^\circ$, $\sin 0^\circ = 0$. Thus, $\overrightarrow{A} \times \overrightarrow{A} = \vec{0}$ This holds regardless of the magnitude of $\overrightarrow{A}$ (as long as it is non-zero).

Given that $|\overrightarrow{A}|$ is a non-zero constant, we analyze each option:

• From the dot product formula, $\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}|^2$.
• Since $|\overrightarrow{A}| \neq 0$, $|\overrightarrow{A}|^2 > 0$.
• Thus, $\overrightarrow{A} \cdot \overrightarrow{A} \neq 0$. Option A is false.

• From the cross product formula, $\overrightarrow{A} \times \overrightarrow{A} = \vec{0}$.
• The zero vector has no direction and its magnitude is zero, so it cannot be negative. Option B is false.

• As derived, $\overrightarrow{A} \times \overrightarrow{A} = \vec{0}$ for any vector $\overrightarrow{A}$.
• This holds true regardless of the magnitude of $\overrightarrow{A}$ (as long as it is defined). Option C is true.

• The cross product yields a vector, not a scalar. Comparing a vector to a scalar (like $> 0$) is mathematically invalid.
• Even if interpreted as the magnitude, $|\overrightarrow{A} \times \overrightarrow{A}| = 0$, which is not greater than zero. Option D is false.

Students often confuse the properties of dot and cross products, leading to the following mistakes:

• Misapplying the dot product: Some assume $\overrightarrow{A} \cdot \overrightarrow{A} = 0$ implies $\overrightarrow{A} = \vec{0}$, forgetting that the dot product of a non-zero vector with itself is always positive.
• Sign confusion in cross product: Students may think $\overrightarrow{A} \times \overrightarrow{A}$ could be positive or negative, not realizing it is always the zero vector. The cross product is anti-commutative ($\overrightarrow{A} \times \overrightarrow{B} = - \overrightarrow{B} \times \overrightarrow{A}$), but this does not apply when the vectors are identical.
• Magnitude vs. vector comparison: Option D incorrectly compares a vector to a scalar. Always remember that cross products yield vectors, not scalars.

Exam Tip: For any vector $\overrightarrow{A}$, memorize these two identities: $\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}|^2$ $\overrightarrow{A} \times \overrightarrow{A} = \vec{0}$ These are fundamental and frequently tested in vector algebra problems.