vectors: help solving part d?

Question

consider a parallelogram ABCD with A(3,-2,-1), B(2,1,3) and C(0,4,1). a-find the coordinates of D b-calculate the area of this parallelogram c-calculate the area of triangle ABD and that of ABC. d-calculate cos BAC and sin BAC

Answer 1

To find the cosine of $BAC$ one can use the geometric definition of the dot product: $v\cdot u=\Vert{v}\Vert\Vert{u}\Vert\cos{\theta}$ where $v,u\in\Bbb R^3$ are vectors and $\theta$ is the angle between them. Substitute $v=AB=B-A$ and $u=AC=C-A$ and use the formula $\Vert{v}\Vert=\sqrt{v_1^2+v_2^2+v_3^2}$ where $v_1,v_2,v_3\in\Bbb R$ are the entries of $v$. Plug in all of the values and rearrange to solve for $\cos{\theta}=\cos{BAC}$. Then use the identity $\sin{\theta}=\cos({\frac{\pi}{2}-\theta})$.