Calculate the area of the parallelogram with adjacent vertices (1,4, −2), (−3,1,6) 𝑦 (1, −2,3)

To calculate the area of a parallelogram with adjacent vertices, we can use the formula:

\text{{Area}} = \left| \vec{u} \times \vec{v} \right|

where \vec{u} and \vec{v} are two vectors formed from the adjacent vertices of the parallelogram.

First, let's find the vectors \vec{u} and \vec{v} using the given vertices.

\vec{u} = \begin{pmatrix} -3 - 1 \ 1 - 4 \ 6 - (-2) \end{pmatrix} = \begin{pmatrix} -4 \ -3 \ 8 \end{pmatrix}

\vec{v} = \begin{pmatrix} 1 - 1 \ -2 - 4 \ 3 - (-2) \end{pmatrix} = \begin{pmatrix} 0 \ -6 \ 5 \end{pmatrix}

Next, let's calculate the cross product of \vec{u} and \vec{v} :

\vec{u} \times \vec{v} = \begin{pmatrix} -4 \ -3 \ 8 \end{pmatrix} \times \begin{pmatrix} 0 \ -6 \ 5 \end{pmatrix} = \begin{pmatrix} (-3) \cdot 5 - 8 \cdot (-6) \ 8 \cdot 0 - (-4) \cdot 5 \ (-4) \cdot (-6) - (-3) \cdot 0 \end{pmatrix} = \begin{pmatrix} 57 \ 20 \ -24 \end{pmatrix}

Now, let's find the magnitude (length) of the cross product vector:

\left| \vec{u} \times \vec{v} \right| = \sqrt{57^2 + 20^2 + (-24)^2} = \sqrt{3249 + 400 + 576} = \sqrt{4225} = 65

Therefore, the area of the parallelogram is 65 square units.


Answer: The area of the parallelogram is 65 square units.