The position of a particle is given by r - -2-k- m and linear momentum as p - 3-4-2k- kg-m-s- The angular momentum of the particle about the origin 0-0-0 is perpendicular to

Angular momentum about the origin (0,0,0) is given by, L =r×p ⇒L = ( 4+4)î ( 2+3)ĵ + (4 6)k̂ ⇒L = ( ĵ 2k̂) kg m^2/s It indicates that the angular momentum v ...

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Standard XIII

Physics

Cross Product

The position ...

Question

The position of a particle is given by $\to r = (^i + 2^j -^k) m$ and linear momentum as $\to p = (3^i + 4^j - 2^k) kg-m/s$ . The angular momentum of the particle about the origin $(0, 0, 0)$ is perpendicular to

x -

axis

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y -

axis

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z -

axis

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x z -

plane

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Solution

The correct option is A $x -$ axis
Angular momentum about the origin $(0, 0, 0)$ is given by,
$\to L = \to r \times \to p$

$\Rightarrow \to L = (- 4 + 4)^i - (- 2 + 3)^j + (4 - 6)^k$
$\Rightarrow \to L = (-^j - 2^k) {kg-m}^{2} /s$
It indicates that the angular momentum vector is in $y z -$ plane, and we know that $y z -$ plane is perpendicular to $x -$ axis.

Hence, the angular momentum $\to L$ is perpendicular to $x -$ axis.

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The position of a particle is given by →r=(^i+2^j−^k) m and linear momentum as →p=(3^i+4^j−2^k) kg-m/s. The angular momentum of the particle about the origin (0,0,0) is perpendicular to

The position of a particle is given by $\to r = (^i + 2^j -^k) m$ and linear momentum as $\to p = (3^i + 4^j - 2^k) kg-m/s$ . The angular momentum of the particle about the origin $(0, 0, 0)$ is perpendicular to