Find the area of the parallelogram that has the vectors as adjacent sides.
$$\mathbf { u } = 2 \mathbf { i } + \mathbf { j } + \mathbf { k } , \mathbf { v } = 5 \mathbf { i } - 5 \mathbf { j } + 5 \mathbf { k }$$
A) 5
B) $\sqrt { 14 }$
C) $\sqrt { 10 }$
D) 4
E) $5 \sqrt { 14 }$

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, $\|\mathbf{u} \times \mathbf{v}\|$. Calculating the cross product:$$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 1 & 1 \ 5 & -5 & 5 \end{vmatrix} = (1\cdot5 - 1\cdot(-5))\mathbf{i} - (2\cdot5 - 1\cdot5)\mathbf{j} + (2\cdot(-5) - 1\cdot5)\mathbf{k} = 10\mathbf{i} + 5\mathbf{j} - 15\mathbf{k}.$$The magnitude of this vector is $\sqrt{10^2 + 5^2 + (-15)^2} = \sqrt{100 + 25 + 225} = \sqrt{350} = 5\sqrt{14}$.

Find the Area of the Parallelogram That Has the Vectors u=2i+j+k,v=5i−5j+5k\mathbf { u } = 2 \mathbf { i } + \mathbf { j } + \mathbf { k } , \mathbf { v } = 5 \mathbf { i } - 5 \mathbf { j } + 5 \mathbf { k }u=2i+j+k,v=5i−5j+5k

Find the Area of the Parallelogram That Has the Vectors $\mathbf { u } = 2 \mathbf { i } + \mathbf { j } + \mathbf { k } , \mathbf { v } = 5 \mathbf { i } - 5 \mathbf { j } + 5 \mathbf { k }$