Answer
The angular momentum about the origin is $~~(-174~kg~m^2/s)~\hat{k}$
Work Step by Step
We can write the general form of the cross-product:
$a\times b = (a_xb_y-b_xa_y)~\hat{k}$
We can express the particle's position in unit-vector notation:
$r = (3.0~m)~\hat{i}+ (8.0~m)~\hat{j}$
We can express the velocity in unit-vector notation:
$v = (5.0~m/s)~\hat{i}-(6.0~m/s)~\hat{j}$
We can find the angular momentum about the origin:
$L = m~(r \times v)$
$L = (3.0~kg)~[(3.0~m)(-6.0~m/s)-(5.0~m/s)(8.0~m)]~\hat{k}$
$L = (3.0~kg)~(-58~m^2/s)~\hat{k}$
$L = (-174~kg~m^2/s)~\hat{k}$
The angular momentum about the origin is $~~(-174~kg~m^2/s)~\hat{k}$
