Find the area of the parallelogram that has the vectors as adjacent sides.
$$\mathbf { u } = \langle 4,2 , - 5 \rangle , \mathbf { v } = \langle - 4,1,2 \rangle$$
A) 14
B) $\sqrt { 41 }$
C) $2 \sqrt { 6 }$
D) 11
E) $3 \sqrt { 41 }$

Question

Quizplus · Accepted Answer

The area of the parallelogram is given by the magnitude of the cross product of the vectors $\mathbf{u}$ and $\mathbf{v}$. The cross product $\mathbf{u} \times \mathbf{v}$ is $\langle 9, 6, 12 \rangle$, and its magnitude is $\sqrt{9^2 + 6^2 + 12^2} = \sqrt{41 \times 9} = 3\sqrt{41}$.

Find the Area of the Parallelogram That Has the Vectors u=⟨4,2,−5⟩,v=⟨−4,1,2⟩\mathbf { u } = \langle 4,2 , - 5 \rangle , \mathbf { v } = \langle - 4,1,2 \rangleu=⟨4,2,−5⟩,v=⟨−4,1,2⟩

Find the Area of the Parallelogram That Has the Vectors $\mathbf { u } = \langle 4,2 , - 5 \rangle , \mathbf { v } = \langle - 4,1,2 \rangle$