Question

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

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Hank

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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.

where

First, let's find the vectors

Next, let's calculate the cross product of

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

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

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

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