Question

Two particles are located at equal distance from origin. The position vectors of those are represented by \( \vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k} \) and \( \vec{B} = 2\hat{i} - 2\hat{j} + 4\hat{k} \), respectively. If both the vectors are at right angle to each other, the value of \( n^{-1} \) is:

Hint:

The result follows from the fact that the vectors are orthogonal and that the distance relations form a solvable system. Understanding the dot product and geometry helps in solving such vector problems.

Answer 1

The dot product of \( \vec{A} \) and \( \vec{B} \) is given by: \[ \vec{A} \cdot \vec{B} = 0 \] This implies: \[ 4 - 6n + 8p = 0 \] Now calculating \( |\vec{A}| \) and \( |\vec{B}| \): \[ |\vec{A}| = \sqrt{2^2 + 3^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \] \[ |\vec{B}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \] Using the formula: \[ |\vec{A}| |\vec{B}| = 4 + 9n^2 + 4 + 4 + 16p^2 = 9n^2 = 16p^2 \] Simplifying, we find: \[ p = \frac{3}{4} n \] \[ 4 - 6n + 6n = 0 \] Thus: \[ n = \frac{1}{3} \]