The position vectors of two $1\ \text{kg}$ particles, (A) and (B), are given by $\vec r_A = (\alpha_1t^2\hat i + \alpha_2t\hat j + \alpha_3t\hat k)\ \text{m}$ and $\vec r_B = (\beta_1t\hat i + \beta_2t^2\hat j + \beta_3t\hat k)\ \text{m}$, respectively; ($\alpha_1 = 1\ \text{m/s}^2$, $\alpha_2 = 3n\ \text{m/s}$, $\alpha_3 = 2\ \text{m/s}$, $\beta_1 = 2\ \text{m/s}$, $\beta_2 = -1\ \text{m/s}^2$, $\beta_3 = 4p\ \text{m/s}$), where $t$ is time, $n$ and $p$ are constants. At $t = 1\ \text{s}$, $|\vec V_A| = |\vec V_B|$ and velocities $\vec V_A$ and $\vec V_B$ of the particles are orthogonal to each other. At $t = 1\ \text{s}$, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $\sqrt{L}\ \text{kg m}^2\text{s}^{-1}$. The value of $L$ is ______.
Numerical value type. Enter your answer.
Answer: 90
At $t = 1\ \text{s}$:
$$\vec V_A = 2\hat i + 3n\hat j + 2\hat k, \qquad \vec V_B = 2\hat i - 2\hat j + 4p\hat k$$
Equal magnitudes: $8 + 9n^2 = 8 + 16p^2 \Rightarrow 3n = \pm4p$.
Orthogonal: $4 - 6n + 8p = 0 \Rightarrow 3n = 2 + 4p$. With $3n = 4p$ this is impossible, so $3n = -4p$, giving $p = -\tfrac{1}{4}$, $n = \tfrac{1}{3}$.
Then $\vec V_A = 2\hat i + \hat j + 2\hat k$, $\vec r_A = \hat i + \hat j + 2\hat k$, $\vec r_B = 2\hat i - \hat j - \hat k$.
Position of A relative to B: $\vec r = \vec r_A - \vec r_B = -\hat i + 2\hat j + 3\hat k$.
$$\vec L = m\,\vec r\times\vec V_A = \begin{vmatrix}\hat i & \hat j & \hat k\\ -1 & 2 & 3\\ 2 & 1 & 2\end{vmatrix} = \hat i + 8\hat j - 5\hat k$$
$$|\vec L|^2 = 1 + 64 + 25 = 90 \Rightarrow L = 90$$
Solution by Sreeraj P, M.Sc Physics