Given vectors

$\vec A = 2\hat i + 3n\hat j + 2\hat k$ $\vec B = 2\hat i - 2\hat j + 4p\hat k$

We are told:

• both particles are at equal distance from origin
• the vectors are perpendicular to each other.

Use equal magnitudes

Since both particles are at equal distance from origin,

$|\vec A| = |\vec B|$

So,

$|\vec A|^2 = |\vec B|^2$

$2^2 + (3n)^2 + 2^2 = 2^2 + (-2)^2 + (4p)^2$

$4 + 9n^2 + 4 = 4 + 4 + 16p^2$

$8 + 9n^2 = 8 + 16p^2$

$9n^2 = 16p^2 \quad \text{...(1)}$

Use perpendicular condition

For right angle,

$\vec A \cdot \vec B = 0$

$2\cdot 2 + (3n)(-2) + 2(4p) = 0$

$4 - 6n + 8p = 0$

$4p - 3n = -2 \quad \text{...(2)}$

Solve equations

From (1):

$9n^2 = 16p^2$

$3n = \pm 4p$

So two cases:

Case 1: $3n = 4p$

Substitute into (2):

$4p - 3n = 0$

But (2) says

$4p - 3n = -2$

Contradiction. So this case is not possible.

Case 2: $3n = -4p$

Then

$4p = -3n$

Put into (2):

$-3n - 3n = -2$

$-6n = -2$

$n = \frac{1}{3}$

Find $n^{-1}$

$n^{-1} = \frac{1}{n} = \frac{1}{1/3} = 3$

Comparison with stored answer

Derived answer = $3$

Stored correct answer = $3$

Hence, they agree.