Question

The linear momentum of a particle $\overrightarrow{P}=3\hat{i}+4\hat{j}-2\hat{k}$ at a point with position vector is given by $\overrightarrow{r}=\hat{i}+2\hat{j}-\hat{k}$ . Then its angular momentum about the origin is perpendicular to,

• A. $x$ -axis
• B. $y$ -axis
• C. $z$ -axis
• D. $yz$ -axis

Answer 1

Answer: (A)

Here, $\overrightarrow{r}=\hat{i}+2\hat{j}-\hat{k},\overrightarrow{p}=3\hat{i}+4\hat{j}-2\hat{k}$
$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ 3& 4& -2\end{array}\right|$
$=\hat{i}\left(-4+4\right)+\hat{j}\left(-3+2\right)+\hat{k}\left(4-6\right)=0\hat{i}-1\hat{j}-2\hat{k}$
$\overrightarrow{L}$ has components along $-y$ axis but it has no component along in the $x-$ axis. The angular momentum $\overrightarrow{L}$ is in $yz$ -plane. $i.e.$ perpendicular to $x-$ axis.