Question

The angular momentum is given by vector L =vector r× vector p where r is position vector
and p is linear momentum of the body. It r=41 +61 -3k and
p=2 i +
4 5k find L and magnitude of L
23​

Answer 1

Given,

$\displaystyle\longrightarrow\sf{\overrightarrow{\sf{r}}=4\hat i+6\hat j-3\hat k}$

$\displaystyle\longrightarrow\sf{\overrightarrow{\sf{p}}=2\hat i+4\hat j+5\hat k}$

Then,

$\displaystyle\longrightarrow\sf{\overrightarrow{\sf{L}}=\overrightarrow{\sf{r}}\times\overrightarrow{\sf{p}}}$

$\displaystyle\longrightarrow\sf{\overrightarrow{\sf{L}}=(4\hat i+6\hat j-3\hat k)\times(2\hat i+4\hat j+5\hat k)}$

$\displaystyle\longrightarrow\sf{\overrightarrow{\sf{L}}=\left|\begin{array}{ccc}\sf{\hat i}&\sf{\hat j}&\sf{\hat k}\\\sf{4}&\sf{6}&\sf{-3}\\\sf{2}&\sf{4}&\sf{5}\end{array}\right|}$

$\displaystyle\longrightarrow\sf{\overrightarrow{\sf{L}}=(6\times5-(-3)\times4)\hat i-(4\times5-(-3)\times2)\hat j+(4\times4-6\times2)\hat k}$

$\displaystyle\longrightarrow\sf{\overrightarrow{\sf{L}}=(30-(-12))\hat i-(20-(-6))\hat j+(16-12)\hat k}$

$\displaystyle\longrightarrow\sf{\overrightarrow{\sf{L}}=(30+12)\hat i-(20+6)\hat j+(16-12)\hat k}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\overrightarrow{\sf{L}}=42\hat i-26\hat j+4\hat k}}}$

And its magnitude is,

$\displaystyle\longrightarrow\sf{\left|\overrightarrow{\sf{L}}\right|=\sqrt{42^2+(-26)^2+4^2}}$

$\displaystyle\longrightarrow\sf{\left|\overrightarrow{\sf{L}}\right|=\sqrt{1764+676+16}}$

$\displaystyle\longrightarrow\sf{\left|\overrightarrow{\sf{L}}\right|=\sqrt{2456}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\left|\overrightarrow{\sf{L}}\right|\simeq49.56\ Js}}}$