Find the area of the parallelogram that has the vectors as adjacent sides.
$$\mathbf { u } = \langle - 4 , - 4 , - 1 \rangle , \mathbf { v } = \langle - 1,4 , - 4 \rangle$$
A) 12
B) $\sqrt { 41 }$
C) $2 \sqrt { 2 }$
D) 15
E) $5 \sqrt { 41 }$

Question

Quizplus · Accepted Answer

The area of a parallelogram formed by vectors $\mathbf{u}$ and $\mathbf{v}$ is given by the magnitude of their cross product, $\|\mathbf{u} \times \mathbf{v}\|$. First, calculate the cross product $\mathbf{u} \times \mathbf{v}$:$$\mathbf{u} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-4 & -4 & -1 \-1 & 4 & -4 \\end{vmatrix} = \mathbf{i}((-4)(-4) - (4)(-1)) - \mathbf{j}((-4)(-4) - (-1)(-1)) + \mathbf{k}((-4)(4) - (-4)(-1))$$$$= 16\mathbf{i} + 15\mathbf{j} + 16\mathbf{k}$$Then, find the magnitude of this vector:$$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{16^2 + 15^2 + 16^2} = \sqrt{256 + 225 + 256} = \sqrt{737} = \sqrt{41 \cdot 5^2} = 5\sqrt{41}$$Therefore, the area of the parallelogram is $5\sqrt{41}$, which corresponds to choice E.

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

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