JEE PYQ: Vector Algebra - Question ID f8c378224232 (JEE Main 2024)

In vector algebra, the angle between two vectors $\vec{A}$ and $\vec{B}$ is determined by their dot product: $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|\,|\vec{B}|}.$ When we need the angle between a vector $\vec{Q}$ and the resultant of two other vectors, we first compute that resultant explicitly, then apply the dot-product formula.

Step 1: Compute the two given vectors
We are given $\vec{A} = 2\vec{Q} + 2\vec{P},\qquad \vec{B} = 2\vec{Q} - 2\vec{P}.$

Step 2: Find the resultant vector
The resultant $\vec{R}$ is simply the vector sum of $\vec{A}$ and $\vec{B}$: $\vec{R} = \vec{A} + \vec{B} = (2\vec{Q} + 2\vec{P}) + (2\vec{Q} - 2\vec{P}) = 4\vec{Q}.$

Step 3: Identify the two vectors whose angle we seek
We want the angle $\theta$ between $\vec{Q}$ and the resultant $\vec{R} = 4\vec{Q}$.

Step 4: Apply the dot-product formula
$\cos \theta = \frac{\vec{Q} \cdot \vec{R}}{|\vec{Q}|\,|\vec{R}|} = \frac{\vec{Q} \cdot (4\vec{Q})}{|\vec{Q}|\,|4\vec{Q}|} = \frac{4\,(\vec{Q}\cdot\vec{Q})}{4\,|\vec{Q}|^2} = \frac{4\,Q^2}{4\,Q^2} = 1.$ Hence $\theta = \cos^{-1}(1) = 0^\circ$.

Conclusion
The angle between $\vec{Q}$ and the resultant is $0^\circ$, so the correct choice is B.

1. Misidentifying the resultant: Students often forget to add the two given vectors and instead try to work with each separately. 2. Overcomplicating the algebra: Expanding dot products prematurely can lead to unnecessary terms. Here, the cancellation of $\vec{P}$ terms is immediate if one adds first. 3. Ignoring scalar multiples: The factor of $4$ in $\vec{R}=4\vec{Q}$ cancels out neatly in the cosine formula, leaving a trivial result.

Tip: Always compute the resultant vector explicitly before attempting to find angles or magnitudes.