The position of a particle is given by r = hati+2 hatj- hatk and its linear momentum is given by p = 3 hati+4 hatj-2 hatk . Then its angular momentum, about the origin is perpendicular to

Question

The position of a particle is given by r = hati+2 hatj- hatk and its linear momentum is given by p = 3 hati+4 hatj-2 hatk . Then its angular momentum, about the origin is perpendicular to (A)  yz- plane (B)  z- axis (C)  y- axis (D)  x- axis

Tardigrade Admin · Accepted Answer

L=r× P=| beginmatrix hati hatj hatk 1 2 -1 3 4 -2 endmatrix | L= hati(-4+4)- hatj(-2+3)+ hatk(4-6) =- hatj-2 hatk L has components along -y axis and -z axis. The angular momentum is in y-z plane ie, perpendicular to x- axis

Q. The position of a particle is given by $r = \hat{i} + 2 \hat{j} - \hat{k}$ and its linear momentum is given by $p = 3 \hat{i} + 4 \hat{j} - 2 \hat{k}$ . Then its angular momentum, about the origin is perpendicular to

$yz -$ plane

$z -$ axis

$y -$ axis

$x -$ axis

$x z -$ plane

$L = r \times P = ∣ ∣ \hat{i} 13 \hat{j} 24 \hat{k} - 1 - 2 ∣ ∣$
$L = \hat{i} (- 4 + 4) - \hat{j} (- 2 + 3) + \hat{k} (4 - 6)$
$= - \hat{j} - 2 \hat{k}$ $L$
has components along -y axis and $- z$ axis. The angular momentum is in y-z plane ie, perpendicular to $x -$ axis

Q. The position of a particle is given by r=i^+2j^​−k^ and its linear momentum is given by p=3i^+4j^​−2k^ . Then its angular momentum, about the origin is perpendicular to

yz− plane

z− axis

y− axis

x− axis

xz− plane

L=r×P=∣∣​i^13​j^​24​k^−1−2​∣∣​ L=i^(−4+4)−j^​(−2+3)+k^(4−6) =−j^​−2k^ L has components along -y axis and −z axis. The angular momentum is in y-z plane ie, perpendicular to x− axis