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

Step 1: Perpendicularity Condition

Vectors are perpendicular: \[ \vec{A} \cdot \vec{B} = 0 \]. Therefore, \[ 4 - 6n + 8p = 0 \tag{1} \]

Step 2: Equal Magnitude Condition

Magnitudes are equal: \[ |\vec{A}| = |\vec{B}| \]. Using the magnitude formula: \[ \sqrt{4 + 9n^2 + 4} = \sqrt{4 + 4 + 16p^2} \]

Step 3: Equation Simplification

Squaring both sides: \[ 4 + 9n^2 + 4 = 4 + 4 + 16p^2 \]. This simplifies to \[ 9n^2 = 16p^2 \], yielding \( p = \pm \frac{3}{4}n \).

Step 4: Substitution into Dot Product Equation

Substitute \( p = \frac{3}{4}n \) into equation (1): \[ 4 - 6n + 8\left(\frac{3}{4}n\right) = 0 \]. This results in \[ 4 - 6n + 6n = 0 \Rightarrow 4 = 0 \], which is a contradiction. Thus, \( p = \frac{3}{4}n \) is not a valid solution.

Step 5: Substitution of Negative Value

Now substitute \( p = -\frac{3}{4}n \): \[ 4 - 6n + 8\left(-\frac{3}{4}n\right) = 0 \]. This leads to \[ 4 - 6n - 6n = 0 \], so \[ 4 - 12n = 0 \Rightarrow n = \frac{1}{3} \].

Final Answer:

$\boxed{n = \dfrac{1}{3}}$