Q152 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
If $\vec{r}$ is a position vector of the body and $\vec{p}$ is linear momentum, $\vec{r}=2\hat{i}+3\hat{j}-3\hat{k}$ and $\vec{p}=2\hat{i}+4\hat{j}-5\hat{k}$.
The angular momentum of the body is
The angular momentum of the body is
- Correct answerA$-3\hat{i}+4\hat{j}+2\hat{k}$
- B$-3\hat{i}-4\hat{j}+2\hat{k}$
- C$-2\hat{i}+3\hat{j}+2\hat{k}$
- D$-\hat{i}-2\hat{j}+\hat{k}$
Solution
The angular momentum $\vec{L}$ of a body is given by the cross product of its position vector $\vec{r}$ and linear momentum $\vec{p}$.
$\vec{L} = \vec{r} \times \vec{p}$
Substituting the given vectors:
$\vec{L} = (2\hat{i} + 3\hat{j} - 3\hat{k}) \times (2\hat{i} + 4\hat{j} - 5\hat{k})$
$\vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -3 \\ 2 & 4 & -5 \end{vmatrix}$
Expanding the determinant:
$\vec{L} = \hat{i}[3(-5) - (-3)(4)] - \hat{j}[2(-5) - (-3)(2)] + \hat{k}[2(4) - 3(2)]$
$\vec{L} = \hat{i}(-15 + 12) - \hat{j}(-10 + 6) + \hat{k}(8 - 6)$
$\vec{L} = -3\hat{i} + 4\hat{j} + 2\hat{k}$
Answer: $-3\hat{i}+4\hat{j}+2\hat{k}$
$\vec{L} = \vec{r} \times \vec{p}$
Substituting the given vectors:
$\vec{L} = (2\hat{i} + 3\hat{j} - 3\hat{k}) \times (2\hat{i} + 4\hat{j} - 5\hat{k})$
$\vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -3 \\ 2 & 4 & -5 \end{vmatrix}$
Expanding the determinant:
$\vec{L} = \hat{i}[3(-5) - (-3)(4)] - \hat{j}[2(-5) - (-3)(2)] + \hat{k}[2(4) - 3(2)]$
$\vec{L} = \hat{i}(-15 + 12) - \hat{j}(-10 + 6) + \hat{k}(8 - 6)$
$\vec{L} = -3\hat{i} + 4\hat{j} + 2\hat{k}$
Answer: $-3\hat{i}+4\hat{j}+2\hat{k}$
Hence, the correct answer is option: 'A'.