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 two vectors $\mathbf{u}$ and $\mathbf{v}$ can be found by calculating the magnitude of their cross product, $\|\mathbf{u} \times \mathbf{v}\|$. The cross product $\mathbf{u} \times \mathbf{v}$ is given by the determinant of the matrix:$$\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-4 & -4 & -1 \-1 & 4 & -4 \\end{vmatrix}= \mathbf{i}(-4(-4) - 4(-1)) - \mathbf{j}(-4(-4) - (-1)(-1)) + \mathbf{k}(-4(4) - (-4)(-1))= \mathbf{i}(16 + 4) - \mathbf{j}(16 - 1) + \mathbf{k}(16 - 4)= \mathbf{i}(20) - \mathbf{j}(15) + \mathbf{k}(12)$$Thus, $\mathbf{u} \times \mathbf{v} = \langle 20, -15, 12 \rangle$. The magnitude of this vector is:$$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{20^2 + (-15)^2 + 12^2} = \sqrt{400 + 225 + 144} = \sqrt{769}$$Since $769 = 41 \times 5^2$, the magnitude simplifies to:$$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{41 \times 25} = 5\sqrt{41}$$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 $\mathbf { u } = \langle - 4 , - 4 , - 1 \rangle , \mathbf { v } = \langle - 1,4 , - 4 \rangle$