A particle has the position vector $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$ and the linear momentum $\vec{p} = 2\hat{i} - \hat{j} + \hat{k}$. Its angular momentum about the origin is}

Question

Rate of cooling of a body is $0.2^\circ \text{C/min}$ when excess temperature is $20^\circ \text{C}$. The proportionality constant $ k $ is:

Answer 1

To find the angular momentum of the particle about the origin, we use the formula:

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{L}$ is the angular momentum, $\vec{r}$ is the position vector, and $\vec{p}$ is the linear momentum.

The position vector is given by:

$\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$

The linear momentum is given by:

$\vec{p} = 2\hat{i} - \hat{j} + \hat{k}$

To calculate the cross product, we set up the determinant:

	$\hat{i}$	$\hat{j}$	$\hat{k}$
Determinant	1	-2	1
	2	-1	1

The cross product is calculated as:

$\vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 2 & -1 & 1 \end{vmatrix}$

Expand the determinant:

$\vec{L} = \hat{i}((-2)(1) - (-1)(1)) - \hat{j}((1)(1) - (2)(1)) + \hat{k}((1)(-1) - (-2)(2))$

$\vec{L} = \hat{i}(-2 + 1) - \hat{j}(1 - 2) + \hat{k}(-1 + 4)$

$\vec{L} = -\hat{i} + \hat{j} + 3\hat{k}$

Therefore, the angular momentum about the origin is:

$-\hat{i} + \hat{j} + 3\hat{k}$

Thus, the correct answer is:

$\boxed{-\hat{i} + \hat{j} + 3\hat{k}}$

Show Hint

	\(\hat{i}\)	\(\hat{j}\)	\(\hat{k}\)
Determinant	1	-2	1
	2	-1	1