JEE PYQ: Vector Algebra - Question ID a90b294abf81 (JEE Main 2024)

Core Formula & Concept:

In vector algebra, two key concepts are used here:

1. Magnitude of a vector: For any vector $\vec{V} = a \hat{i} + b \hat{j}$, its magnitude is given by $|\vec{V}| = \sqrt{a^2 + b^2}.$
2. Parallel vectors: Two vectors are parallel if and only if one is a scalar multiple of the other. If $\vec{U}$ is parallel to $\vec{B}$, then $\vec{U} = k \vec{B}$ for some scalar $k > 0$ (since the vector lies in the first quadrant).

Step-by-Step Derivation:

Step 1: Compute the magnitude of $\vec{A}$.

Given $\vec{A} = 3 \hat{i} + 4 \hat{j}$, its magnitude is $|\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$

Step 2: Express the desired vector in terms of $\vec{B}$.

The desired vector $\vec{C}$ has the same magnitude as $\vec{A}$ and is parallel to $\vec{B} = 4 \hat{i} + 3 \hat{j}$. Since $\vec{C}$ is parallel to $\vec{B}$, we can write $\vec{C} = k \vec{B} = k(4 \hat{i} + 3 \hat{j}) = 4k \hat{i} + 3k \hat{j},$ where $k > 0$ (because $\vec{C}$ lies in the first quadrant).

Step 3: Use the magnitude condition to find $k$.

The magnitude of $\vec{C}$ must equal $|\vec{A}| = 5$: $|\vec{C}| = \sqrt{(4k)^2 + (3k)^2} = \sqrt{16k^2 + 9k^2} = \sqrt{25k^2} = 5k.$ Set this equal to 5: $5k = 5 \implies k = 1.$

Step 4: Identify the components of $\vec{C}$.

Substituting $k = 1$ into $\vec{C}$ gives $\vec{C} = 4 \hat{i} + 3 \hat{j}.$ However, the problem states that the $y$-component of $\vec{C}$ is 3, which matches the above expression. The $x$-component is therefore $4$.

Verification:

The problem specifies that the $x$ and $y$ components of the vector in the first quadrant are $x$ and 3 respectively. From the above, $x = 4$. Common Traps & Exam Tip:

Students often confuse the direction of parallel vectors, mistakenly assuming $k$ can be negative. Since the vector lies in the first quadrant, $k$ must be positive. Additionally, some students misapply the magnitude formula by forgetting to square the components or incorrectly solving for $k$. Always double-check the quadrant condition to ensure the correct sign of $k$.