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:
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Approach Solution - 1
Step 1: Condition for perpendicularity
Since the vectors are perpendicular, \[ \vec{A} \cdot \vec{B} = 0 \] Hence, \[ 4 - 6n + 8p = 0 \tag{1} \]
Step 2: Condition for equal magnitudes
Given that the magnitudes are equal: \[ |\vec{A}| = |\vec{B}| \] Using the magnitude formula: \[ \sqrt{4 + 9n^2 + 4} = \sqrt{4 + 4 + 16p^2} \]
Step 3: Simplify the equation
Squaring both sides: \[ 4 + 9n^2 + 4 = 4 + 4 + 16p^2 \] \[ 9n^2 = 16p^2 \] \[ p = \pm \frac{3}{4}n \]
Step 4: Substitute into the dot product equation
Substitute \( p = \frac{3}{4}n \) into (1): \[ 4 - 6n + 8\left(\frac{3}{4}n\right) = 0 \] \[ 4 - 6n + 6n = 0 \Rightarrow 4 = 0 \] This is a contradiction, so \( p = \frac{3}{4}n \) is not valid.
Step 5: Substitute the negative value
Now take \( p = -\frac{3}{4}n \): \[ 4 - 6n + 8\left(-\frac{3}{4}n\right) = 0 \] \[ 4 - 6n - 6n = 0 \] \[ 4 - 12n = 0 \Rightarrow n = \frac{1}{3} \]
Final Answer:
$\boxed{n = \dfrac{1}{3}}$
Approach Solution -2
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: \( n = \frac{1}{3} \)
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