The position of a particle is given by : r = î + 2ĵ - k̂ and momentum P = 3î + 4ĵ - 2k̂ . The angular momentum is perpendicular to

The correct option is X-axis.

To find which axis the angular momentum of the particle is perpendicular to, we need to calculate the angular momentum vector ($\stackrel{��}{L}$) and check its components.

The angular momentum $\overrightarrow{L}$ of a particle is defined by the cross product of its position vector $\overrightarrow{r}$ and its linear momentum vector $\overrightarrow{p}$:
$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Given vectors:
$\overrightarrow{r}=1\hat{i}+2\hat{j}-1\hat{k}$
$\overrightarrow{p}=3\hat{i}+4\hat{j}-2\hat{k}$

We compute the cross product using a determinant:

Expanding the determinant by the first row:

Now, calculate each component term by term:

For the X-component ($L_x$):
$L_x=-4-(-4)=-4+4=0$

For the Y-component ($L_y$):
$L_y=-[-2-(-3\left)\right]=-[-2+3]=-1$

For the Z-component ($L_z$):
$L_z=4-6=-2$

Therefore, the angular momentum vector is:

A vector is perpendicular to a coordinate axis if its component along that axis is zero. Since the X-component ($L_x$) is 0, the angular momentum vector lies entirely in the Y-Z plane.

Taking the dot product of $\overrightarrow{L}$ with the unit vector along the X-axis ($\hat{i}$):
$\overrightarrow{L}·\hat{i}=(0\hat{i}-\hat{j}-2\hat{k})·\hat{i}=0$

Since the dot product is zero, the angular momentum is perpendicular to the X-axis.