The position vectors of two 1 kg particles, (A) and (B), are given by ${\overrightarrow{\mathrm{r}}}_{\mathrm{A}}=\left({\mathrm{\alpha }}_1{\mathrm{t}}^2\widehat{\mathrm{i}}+{\mathrm{\alpha }}_2\mathrm{t}\widehat{\mathrm{j}}+{\mathrm{\alpha }}_3\mathrm{t}\widehat{\mathrm{k}}\right) \mathrm{m}$ and ${\overrightarrow{\mathrm{r}}}_{\mathrm{B}}=\left({\mathrm{\beta }}_1\mathrm{t}\widehat{\mathrm{i}}+{\mathrm{\beta }}_2{\mathrm{t}}^2\widehat{\mathrm{j}}+{\mathrm{\beta }}_3\mathrm{t}\widehat{\mathrm{k}}\right) \mathrm{m}$ respectively $({\mathrm{\alpha }}_1=1 \mathrm{m}/{\mathrm{s}}^2,{\mathrm{\alpha }}_2=3n \mathrm{m}/\mathrm{s},{\mathrm{\alpha }}_3=2 \mathrm{m}/\mathrm{s},{\mathrm{\beta }}_1=2 \mathrm{m}/\mathrm{s},$ ${\beta }_2=-1 \mathrm{m}/{\mathrm{s}}^2,{\beta }_3=4p \mathrm{m}/\mathrm{s}),$ where t is time, $n$ and $p$ are constants. At $\mathrm{t}=1\mathrm{s},\left|{\overrightarrow{\mathrm{V}}}_{\mathrm{A}}\right|=\left|{\overrightarrow{\mathrm{V}}}_{\mathrm{B}}\right|$ and velocities ${\overrightarrow{\mathrm{V}}}_{\mathrm{A}}$ and ${\overrightarrow{\mathrm{V}}}_{\mathrm{B}}$ of the particles are orthogonal to each other. At t = 1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is$\sqrt{L} {\mathrm{kgm}}^2{\mathrm{s}}^{-1}$. The value of $L$ is _____. 

We are given the position vectors of two particles A and B as functions of time:

$$\mathbf{r}_A = \alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 t \hat{k}$$

 $$\mathbf{r}_B = \beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k}$$

Given values:
$\alpha_1 = 1$, $\alpha_2 = 3n$, $\alpha_3 = 2$
$\beta_1 = 2$, $\beta_2 = -1$, $\beta_3 = 4p$

At $t = 1s$, the velocity vectors are:

$$\mathbf{V}_A = 2 \hat{i} + 3n \hat{j} + 2 \hat{k}$$

 $$\mathbf{V}_B = 2 \hat{i} - 2 \hat{j} + 4p \hat{k}$$

Step 1: Apply Given Conditions

Condition 1: Orthogonality of Velocities

$$\mathbf{V}_A \cdot \mathbf{V}_B = 0$$

 $$(2 \times 2) + (3n \times -2) + (2 \times 4p) = 0$$ $$4 - 6n + 8p = 0$$ $$6n = 8p + 4$$

Condition 2: Equal Speeds

$$|\mathbf{V}_A| = |\mathbf{V}_B|$$

 $$4 + 9n^2 + 4 = 4 + 4 + 16p^2$$ 

$$8 + 9n^2 = 8 + 16p^2$$

 $$9n^2 = 16p^2$$

 $$\frac{n^2}{p^2} = \frac{16}{9} \quad \Rightarrow \quad \frac{n}{p} = \pm \frac{4}{3}$$

Step 2: Solve for $n$ and $p$

Case 1: $n/p = 4/3$

No valid solution.

Case 2: $n/p = -4/3$

$$n = -\frac{4}{3} p$$

Substituting into $6n = 8p + 4$:

$$6 \times \left(-\frac{4}{3} p\right) = 8p + 4$$

 $$-8p = 8p + 4$$ $$-16p = 4$$ $$p = -\frac{1}{4}$$ $$n = \frac{1}{3}$$

Step 3: Compute Angular Momentum

Relative Position of A w.r.t. B

$$\mathbf{r}_A - \mathbf{r}_B = (-1) \hat{i} + (3n + 1) \hat{j} + (2 - 4p) \hat{k}$$

Substituting $n = \frac{1}{3}$ and $p = -\frac{1}{4}$:

$$3n + 1 = 2, \quad 2 - 4p = 3$$

 $$\mathbf{r}_A - \mathbf{r}_B = -\hat{i} + 2 \hat{j} + 3 \hat{k}$$

Velocity of A

$$\mathbf{V}_A = 2 \hat{i} + 1 \hat{j} + 2 \hat{k}$$

Compute Angular Momentum,  $\mathit{L}=m({\mathbf{r}}_A-{\mathbf{r}}_B)\times {\mathbf{V}}_A$

$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 2 & 1 & 2 \end{vmatrix}$$

Expanding:

$$\hat{i} (2 \times 2 - 3 \times 1) - \hat{j} (-1 \times 2 - 3 \times 2) + \hat{k} (-1 \times 1 - 2 \times 2)$$

 $$\hat{i} (1) - \hat{j} (-8) + \hat{k} (-5)$$ 

$$\mathbf{L} = 1 \hat{i} + 8 \hat{j} - 5 \hat{k}$$

Magnitude of Angular Momentum

$$L = \sqrt{1^2 + 8^2 + (-5)^2}$$

 $$L = \sqrt{1 + 64 + 25} = \sqrt{90} = 3\sqrt{10}$$