Find the area of the parallelogram whose vertices are 0, a, b, and a+b, where a and b are the vectors defined in part (a)
part (a): Let \({a} = \begin{pmatrix} \phantom -2 \\ -6 \\ \phantom -5 \end{pmatrix} \quad \text{and} \quad {b} = \begin{pmatrix} -1 \\ -2 \\ \phantom -0 \end{pmatrix}.\)find a*b and the answer was 10
+30 \[\sin^2 θ = 1 - \cos^2 θ = 1 - \frac{4}{13} = \frac{9}{13}\]
every angle in a parallelogram is between 0 and π radians, so θ must be positive, so
\[\sin θ = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}\]
area of parallelogram is
\[\|{a}\| \|{b}\| \sin \theta = \sqrt{65} \cdot \sqrt{5} \cdot \frac{3}{\sqrt{13}} = \boxed{15}\]
