Finding Torque using Vector Products 
Let $B \space$be the point at which the force acts.
$$
T= |\underline r \space\times \underline F|
$$
$$
AB =\underline r
$$
$$
AB = 1.5\underline i \space-1\underline j
$$
I'm not quite sure how to find the Force vector, any help would be appreciated.
 A: The general component-wise formula for a cross product is a bit more complicated than I like, 
making computational errors all too likely.
But in this case it is possible to work with much simpler pieces.
A convenient fact about the cross product is that it distributes over the components of its input vectors like multiplication over addition.
That is, we can consider each possible pair of components of the two vectors, work out the cross product of that pair, and add all the results to get the final answer.
For this problem the input vectors have only $\mathbf i$ and $\mathbf j$ components.
Since $\mathbf v \times \mathbf v = 0$ for any vector $\mathbf v,$
the only component pairs that will contribute to the answer are the
$\mathbf i$ component of the moment arm $\mathbf r$ paired with the $\mathbf j$ component of the force $\mathbf F$
and the $\mathbf j$ component of the moment arm $\mathbf r$ paired with the $\mathbf i$ component of the force $\mathbf F.$
That is, the whole problem comes down to a combination of effects that you can treat as two independent sources of torque: the force component $20\mathbf i$ acting perpendicular to the moment arm $-1\mathbf j$
(resulting in a product with magnitude $20$)
and the force component $20\sqrt3\mathbf j$
acting perpendicular to the moment arm $1.5\mathbf i$
(resulting in a product with magnitude $30\sqrt3$).
In order to get the sign of each component correct, we can carefully distinguish between which of these torques is a scalar multiple of the cross product $\mathbf i\times\mathbf j$ and which is a scalar multiple of $\mathbf j\times\mathbf i,$
and observe how the signs of these products interact with the signs of the scalars,
but another way (which I think is easier and less error-prone in this case)
is to observe that each of these components of torque is pulling its moment arm in a counterclockwise direction.
Hence we can simply add the magnitudes of the two component products
to get the total counterclockwise torque.
In other words, the magnitude of the final answer is simply
$$
1(20) + 1.5(20\sqrt3) \approx 20 + 30(1.732) = 20 + 51.96 = 71.96.
$$
A: Indeed your approach is quite slow and cumbersome, which sometimes causes errors.
A definition of cross product is
$r\times F=||r||\cdot||F||\sin\theta \cdot\hat{n}$
where $\theta$ is the angle between the vectors, and $\hat{n}$ is a unit vector perpendicular to the plane of the original vectors(direction determined by right hand rule).
Obviously, $\theta=\tan^{-1}(1.5)+\frac\pi6$.
Hence, the magnitude of the torque is
$$\sqrt{1.5^2+1^2}\cdot40\cdot\sin\theta=71.96 \text{Nm}$$
If you really want to stick with the cross product of Cartesian vectors, notice that
$$(ai+bj)\times(ci+dj)=(ad-bc)k$$where $k$ is the unit vector pointing out of the paper.
