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

Given the condition that vectors \( \vec{A} \cdot \vec{B} = 0 \), we start by applying the dot product:

\( 4 - 6n + 8p = 0 \)

It is also given that the magnitudes of the vectors are equal: \( |\vec{A}| = |\vec{B}| \)

Using the magnitude formula:

\( \sqrt{4 + 9n^2 + 4} = \sqrt{4 + 4 + 16p^2} \)

Squaring both sides and simplifying:

\( 4 + 9n^2 + 4 = 4 + 4 + 16p^2 \)

\( 9n^2 = 16p^2 \)

Solving for \( p \):

\( p = \pm \frac{3}{4}n \)

Substitute \( p = \frac{3}{4}n \) back into the dot product equation:

\( 4 - 6n + 8 \cdot \frac{3}{4}n = 0 \)

\( 4 - 6n + 6n = 0 \)

\( 4 = 0 \) → contradiction? (Note: If we substitute with the opposite sign \( p = -\frac{3}{4}n \), we get:)

\( 4 - 6n - 6n = 0 \Rightarrow 4 - 12n = 0 \Rightarrow n = \frac{1}{3} \)

Final Answer: \( n = \frac{1}{3} \)