QJEEM25Phf3f6: JEE Main 2025 Physics - Class 11 JEEM25Phf3f6 Solution: Two particles are located at equal distance from origin. The…

We are given two conditions:

The particles are at the same distance from the origin. This means that the magnitudes of their position vectors are equal.

The vectors are perpendicular (at right angles) to each other. This means their dot product is zero.

Let’s work through these step-by-step.

● The vectors are given by:

$\vec{A} = 2\hat{i} + 3n \hat{j} + 2\hat{k},$

$\vec{B} = 2\hat{i} -2\hat{j} + 4p \hat{k}.$

● Since the vectors are perpendicular, their dot product must be zero:

$\vec{A} \cdot \vec{B} = (2)(2) + (3n)(-2) + (2)(4p) = 4 - 6n + 8p = 0.$

This can be rearranged to:

$8p = 6n - 4 \quad \Longrightarrow \quad p = \frac{3n - 2}{4} \quad \text{(Equation 1)}.$

● Since the distances from the origin are equal, the magnitudes of the vectors must be equal. Compute the squares of the magnitudes:

For $\vec{A}: \quad |\vec{A}|^2 = 2^2 + (3n)^2 + 2^2 = 4 + 9n^2 + 4 = 8 + 9n^2.$

For $\vec{B}: \quad |\vec{B}|^2 = 2^2 + (-2)^2 + (4p)^2 = 4 + 4 + 16p^2 = 8 + 16p^2.$

Setting them equal:

$8 + 9n^2 = 8 + 16p^2 \quad \Longrightarrow \quad 9n^2 = 16p^2.$

This simplifies to:

$p^2 = \frac{9n^2}{16} \quad \Longrightarrow \quad p = \pm \frac{3n}{4} \quad \text{(Equation 2)}.$

● Now equate the two expressions for $p$ from Equation 1 and Equation 2.

Case 1: Assume $p = \frac{3n}{4}.$

Then,

$\frac{3n - 2}{4} = \frac{3n}{4} \quad \Longrightarrow \quad 3n - 2 = 3n \quad \Longrightarrow \quad -2 = 0,$

which is a contradiction.

Case 2: Assume $p = -\frac{3n}{4}.$

Then,

$\frac{3n - 2}{4} = -\frac{3n}{4} \quad \Longrightarrow \quad 3n - 2 = -3n.$

Simplify by adding $3n$ to both sides:

$6n - 2 = 0 \quad \Longrightarrow \quad 6n = 2 \quad \Longrightarrow \quad n = \frac{1}{3}.$

● Since we found $n = \frac{1}{3},$ its reciprocal is:

$n^{-1} = 3.$

Thus, the value of $n^{-1}$ is $3.$