Find the area of the parallelogram that has the vectors as adjacent sides.
$$\mathbf { u } = - 3 \mathbf { i } - 2 \mathbf { j } + 5 \mathbf { k } , \mathbf { v } = \mathbf { i } - 2 \mathbf { j } + \mathbf { k }$$
A) 1
B) $\sqrt { 3 }$
C) $\sqrt { 6 }$
D) 16
E) $8 \sqrt { 3 }$

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:$$\mathbf{u} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-3 & -2 & 5 \1 & -2 & 1 \\end{vmatrix}= \mathbf{i}(-2-(-10)) - \mathbf{j}(-3-5) + \mathbf{k}((-3)(-2)-(-2)) = 8\mathbf{i} + 8\mathbf{j} + 4\mathbf{k}.$$Then, find the magnitude of this vector:$$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12.$$It appears there was a mistake in the calculation or the options provided. Given the calculation above, the correct magnitude of the cross product is 12, which is not listed among the options. The correct approach to find the area involves calculating the cross product and its magnitude, but the discrepancy suggests a reevaluation of the calculation or the options might be necessary.

Find the Area of the Parallelogram That Has the Vectors u=−3i−2j+5k,v=i−2j+k\mathbf { u } = - 3 \mathbf { i } - 2 \mathbf { j } + 5 \mathbf { k } , \mathbf { v } = \mathbf { i } - 2 \mathbf { j } + \mathbf { k }u=−3i−2j+5k,v=i−2j+k

Find the Area of the Parallelogram That Has the Vectors $\mathbf { u } = - 3 \mathbf { i } - 2 \mathbf { j } + 5 \mathbf { k } , \mathbf { v } = \mathbf { i } - 2 \mathbf { j } + \mathbf { k }$