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(B) Given that the particles are at an equal distance from the origin,their magnitudes are equal: $|\overrightarrow{A}| = |\overrightarrow{B}|$.$|\ove

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given that the particles are at an equal distance from the origin,their magnitudes are equal: $|\overrightarrow{A}| = |\overrightarrow{B}|$.$|\overrightarrow{A}|^2 = |\overrightarrow{B}|^2 \implies 2^2 + (3n)^2 + 2^2 = 2^2 + (-2)^2 + (4p)^2$.$4 + 9n^2 + 4 = 4 + 4 + 16p^2 \implies 8 + 9n^2 = 8 + 16p^2 \implies 9n^2 = 16p^2$.Taking the square root,$3n = \pm 4p$,so $p = \pm \frac{3n}{4}$.Since the vectors are at a right angle,their dot product is zero: $\overrightarrow{A} \cdot \overrightarrow{B} = 0$.$(2)(2) + (3n)(-2) + (2)(4p) = 0 \implies 4 - 6n + 8p = 0$.Case $1$: Substitute $p = \frac{3n}{4}$ into the dot product equation: $4 - 6n + 8(\frac{3n}{4}) = 0 \implies 4 - 6n + 6n = 0 \implies 4 = 0$ (Impossible).Case $2$: Substitute $p = -\frac{3n}{4}$ into the dot product equation: $4 - 6n + 8(-\frac{3n}{4}) = 0 \implies 4 - 6n - 6n = 0 \implies 12n = 4 \implies n = \frac{1}{3}$.Therefore,$n^{-1} = \frac{1}{n} = 3$.

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