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

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## Answer to a math question Calculate the area of the parallelogram with adjacent vertices $1,4, −2$, $−3,1,6$ 𝑦 $1, −2,3$

Hank
4.8
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.

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