JEE PYQ: Vector Algebra - Question ID ffe9ebd89b65 (JEE Main 2025)

In vector algebra, two vectors \(\vec{A}\) and \(\vec{B}\) are said to be perpendicular (at right angles) if and only if their dot product is zero. The dot product is defined as: $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$ where \(\theta\) is the angle between the vectors. When \(\theta = 90^\circ\), \(\cos 90^\circ = 0\), so: $\vec{A} \cdot \vec{B} = 0$ Additionally, the problem states that both particles are located at equal distances from the origin. This implies that the magnitudes of \(\vec{A}\) and \(\vec{B}\) are equal: $|\vec{A}| = |\vec{B}|$

Given the position vectors: $\vec{A} = 2 \hat{i} + 3n \hat{j} + 2 \hat{k}$ $\vec{B} = 2 \hat{i} - 2 \hat{j} + 4p \hat{k}$ Step 1: Use the perpendicularity condition
Since \(\vec{A}\) and \(\vec{B}\) are perpendicular: $\vec{A} \cdot \vec{B} = 0$ Compute the dot product: $(2)(2) + (3n)(-2) + (2)(4p) = 0$ Simplify: $4 - 6n + 8p = 0 \quad \text{(Equation 1)}$ Step 2: Use the equal magnitude condition
Since \(|\vec{A}| = |\vec{B}|\): $\sqrt{2^2 + (3n)^2 + 2^2} = \sqrt{2^2 + (-2)^2 + (4p)^2}$ Square both sides to eliminate the square roots: $4 + 9n^2 + 4 = 4 + 4 + 16p^2$ Simplify: $8 + 9n^2 = 8 + 16p^2$ Subtract 8 from both sides: $9n^2 = 16p^2$ Take the square root of both sides (considering positive values for simplicity, as \(n\) and \(p\) are likely real numbers): $3n = 4p \quad \Rightarrow \quad p = \frac{3n}{4} \quad \text{(Equation 2)}$ Step 3: Substitute \(p\) from Equation 2 into Equation 1
From Equation 1: $4 - 6n + 8p = 0$ Substitute \(p = \frac{3n}{4}\): $4 - 6n + 8 \left( \frac{3n}{4} \right) = 0$ Simplify: $4 - 6n + 6n = 0$ This simplifies to: $4 = 0$ This is a contradiction, which means our assumption that both \(n\) and \(p\) are positive may not hold. Let’s reconsider the square root step in Step 2. The correct approach is to consider both positive and negative roots: $3n = \pm 4p$ Thus, two cases arise: 1. \(p = \frac{3n}{4}\) 2. \(p = -\frac{3n}{4}\) Case 1: \(p = \frac{3n}{4}\)
Substituting into Equation 1: $4 - 6n + 8 \left( \frac{3n}{4} \right) = 0 \quad \Rightarrow \quad 4 - 6n + 6n = 0 \quad \Rightarrow \quad 4 = 0$ This is invalid, so this case is discarded. Case 2: \(p = -\frac{3n}{4}\)
Substituting into Equation 1: $4 - 6n + 8 \left( -\frac{3n}{4} \right) = 0$ Simplify: $4 - 6n - 6n = 0 \quad \Rightarrow \quad 4 - 12n = 0$ Solve for \(n\): $12n = 4 \quad \Rightarrow \quad n = \frac{4}{12} = \frac{1}{3}$ Step 4: Find \(n^{-1}\)
Given \(n = \frac{1}{3}\), the reciprocal is: $n^{-1} = 3$ Thus, the correct answer is \(3\).

1. Ignoring the equal magnitude condition: Many students only use the perpendicularity condition (\(\vec{A} \cdot \vec{B} = 0\)) and forget that the magnitudes are equal. This leads to an underdetermined system with infinite solutions. 2. Sign errors in the square root step: When solving \(9n^2 = 16p^2\), students often consider only the positive root (\(3n = 4p\)) and miss the negative root (\(3n = -4p\)), which is crucial for obtaining the correct solution. 3. Algebraic mistakes: Simple arithmetic errors (e.g., \(8 \times \frac{3n}{4} = 6n\) instead of \(6n\)) can lead to incorrect conclusions. Always double-check calculations. 4. Assuming \(n\) and \(p\) are positive: The problem does not specify the signs of \(n\) and \(p\), so both positive and negative cases must be considered.

Exam Tip: When dealing with vector problems involving perpendicularity and equal magnitudes, always: 1. Write down both conditions explicitly. 2. Solve the system of equations carefully, considering all possible cases (especially signs). 3. Verify the solution by plugging it back into the original conditions.