JEE PYQ: Vector Algebra - Question ID 739104427286 (JEE Main 2024)

In vector algebra, the resultant of two vectors $\vec{A}$ and $\vec{B}$ is given by their vector sum: $\vec{R} = \vec{A} + \vec{B}$ The magnitude of the resultant can be computed using the law of cosines: $|\vec{R}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta$ where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.

A key condition in this problem is that the resultant $\vec{R}$ is perpendicular to $\vec{A}$. When two vectors are perpendicular, their dot product vanishes: $\vec{R} \cdot \vec{A} = 0$ Expanding $\vec{R} = \vec{A} + \vec{B}$ gives $(\vec{A} + \vec{B}) \cdot \vec{A} = 0$ which simplifies to $|\vec{A}|^2 + \vec{B} \cdot \vec{A} = 0$ or equivalently $\vec{B} \cdot \vec{A} = -|\vec{A}|^2.$

The problem also states that the magnitude of $\vec{R}$ is half that of $\vec{B}$: $|\vec{R}| = \tfrac12|\vec{B}|.$

Step 1: Use the perpendicularity condition.
Since $\vec{R}\perp\vec{A}$, $\vec{R}\cdot\vec{A} = 0.$ Substitute $\vec{R} = \vec{A} + \vec{B}$: $(\vec{A} + \vec{B})\cdot\vec{A} = 0 \quad\Longrightarrow\quad |\vec{A}|^2 + \vec{B}\cdot\vec{A} = 0.$ Therefore $\vec{B}\cdot\vec{A} = -|\vec{A}|^2.$ But $\vec{B}\cdot\vec{A} = |\vec{A}||\vec{B}|\cos\theta$, so $|\vec{A}||\vec{B}|\cos\theta = -|\vec{A}|^2 \quad\Longrightarrow\quad |\vec{B}|\cos\theta = -|\vec{A}|.$

Step 2: Use the magnitude condition.
Given $|\vec{R}| = \tfrac12|\vec{B}|$, square both sides: $|\vec{R}|^2 = \tfrac14|\vec{B}|^2.$ By the law of cosines, $|\vec{R}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta.$ Substitute $|\vec{R}|^2 = \tfrac14|\vec{B}|^2$ and $\cos\theta = -\tfrac{|\vec{A}|}{|\vec{B}|}$ from Step 1: $\tfrac14|\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\Bigl(-\tfrac{|\vec{A}|}{|\vec{B}|}\Bigr) = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}|^2 = |\vec{B}|^2 - |\vec{A}|^2.$ Rearrange: $\tfrac14|\vec{B}|^2 = |\vec{B}|^2 - |\vec{A}|^2 \quad\Longrightarrow\quad |\vec{A}|^2 = \tfrac34|\vec{B}|^2 \quad\Longrightarrow\quad |\vec{A}| = \tfrac{\sqrt3}{2}|\vec{B}|.$

Step 3: Find the angle $\theta$.
From Step 1 we have $\cos\theta = -\frac{|\vec{A}|}{|\vec{B}|} = -\frac{\tfrac{\sqrt3}{2}|\vec{B}|}{|\vec{B}|} = -\frac{\sqrt3}{2}.$ The angle whose cosine is $-\tfrac{\sqrt3}{2}$ is $150^\circ$.

1. Sign error in dot product: Many students forget the negative sign when using $\vec{R}\cdot\vec{A}=0$ and write $\vec{B}\cdot\vec{A}=+|\vec{A}|^2$ instead of $-|\vec{A}|^2$. 2. Magnitude substitution: It is easy to mis-substitute $|\vec{R}|^2$ into the law of cosines. Always double-check that you replace $|\vec{R}|^2$ by $\tfrac14|\vec{B}|^2$ and not by $\tfrac12|\vec{B}|^2$. 3. Angle quadrant: The cosine is negative, so the angle must lie in the second quadrant. The principal value is $150^\circ$, not $30^\circ$.