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

Guest Apr 2, 2022

#1**+2 **

\[\sin^2 θ = 1 - \cos^2 θ = 1 - \frac{4}{13} = \frac{9}{13}\]

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

\[\|{a}\| \|{b}\| \sin \theta = \sqrt{65} \cdot \sqrt{5} \cdot \frac{3}{\sqrt{13}} = \boxed{15}\]

justchillin Jun 25, 2022