Question

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

Answer 1

3

Answer 2

Given vectors are $\vec{A} = 2\hat{i}+3n\hat{j}+2\hat{k}$ and $\vec{B}=2\hat{i}-2\hat{j}+4p\hat{k}$.

Condition 1: The particles are located at equal distance from the origin. This means the magnitudes of the position vectors are equal: $|\vec{A}| = |\vec{B}|$. $|\vec{A}|^2 = 2^2 + (3n)^2 + 2^2 = 4 + 9n^2 + 4 = 8 + 9n^2$ $|\vec{B}|^2 = 2^2 + (-2)^2 + (4p)^2 = 4 + 4 + 16p^2 = 8 + 16p^2$ Since $|\vec{A}| = |\vec{B}|$, we have $|\vec{A}|^2 = |\vec{B}|^2$. $8 + 9n^2 = 8 + 16p^2$ $9n^2 = 16p^2$ (Equation 1)

Condition 2: The two vectors are at right angle to each other. This means their dot product is zero: $\vec{A} \cdot \vec{B} = 0$. $\vec{A} \cdot \vec{B} = (2)(2) + (3n)(-2) + (2)(4p)$ $\vec{A} \cdot \vec{B} = 4 - 6n + 8p$ Setting the dot product to zero: $4 - 6n + 8p = 0$ Dividing the equation by 2: $2 - 3n + 4p = 0$ $3n - 4p = 2$ (Equation 2)

Now we need to solve the system of equations (1) and (2) for $n$ and $p$. From Equation 1, $9n^2 = 16p^2$, which can be written as $9n^2 - 16p^2 = 0$. This is a difference of squares: $(3n)^2 - (4p)^2 = 0$. Factoring, we get $(3n - 4p)(3n + 4p) = 0$. This implies either $3n - 4p = 0$ or $3n + 4p = 0$.

We also have Equation 2: $3n - 4p = 2$.

Case 1: $3n - 4p = 0$. If we substitute this into Equation 2, we get $0 = 2$, which is a contradiction. So, this case is not possible.

Case 2: $3n + 4p = 0$. This must be the valid case. We now have a system of two linear equations:

1. $3n + 4p = 0$
2. $3n - 4p = 2$

We can solve this system by adding the two equations: $(3n + 4p) + (3n - 4p) = 0 + 2$ $6n = 2$ $n = \frac{2}{6} = \frac{1}{3}$

Now substitute the value of $n$ into either equation to find $p$. Using the first equation $3n + 4p = 0$: $3\left(\frac{1}{3}\right) + 4p = 0$ $1 + 4p = 0$ $4p = -1$ $p = -\frac{1}{4}$

We have found the values $n = \frac{1}{3}$ and $p = -\frac{1}{4}$. The question asks for the value of $n^{-1}$. $n^{-1} = \left(\frac{1}{3}\right)^{-1} = 3$.