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# Linear Algebra

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The rectangle ABCD has vertices at A = (1, 2, 3), B = (3, 6, −2), and C = (0, 2, −6). Determine the coordinates of vertex D.

The answer according to the book is:  D = (-2, -2, -1)

Jan 23, 2020

#1
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Okay, I think I see where the  (-2, -2, -1)  came from, but I am still confused about this question and I want to see what others have to say.

Jan 23, 2020
edited by hectictar  Jan 23, 2020
#2
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The rectangle ABCD has vertices at A = (1, 2, 3), B = (3, 6, −2), and C = (0, 2, −6). Determine the coordinates of vertex D.

Jan 23, 2020
#3
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The rectangle ABCD has vertices at A = (1, 2, 3), B = (3, 6, −2), and C = (0, 2, −6).

Determine the coordinates of vertex D.

$$\begin{array}{|rcll|} \hline \mathbf{\vec{B}-\vec{A}} &=& \mathbf{\vec{C}-\vec{D}} \\\\ \left( \begin{array}{r} 3 \\ 6 \\ -2 \end{array}\right) - \left( \begin{array}{r} 1 \\ 2 \\ 3 \end{array}\right) &=& \left( \begin{array}{r} 0 \\ 2 \\ -6 \end{array}\right) - \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) \\\\ \left( \begin{array}{r} 3-1 \\ 6-2 \\ -2-3 \end{array}\right) &=& \left( \begin{array}{r} 0 \\ 2 \\ -6 \end{array}\right) - \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) \\\\ \left( \begin{array}{r} 2 \\ 4 \\ -5 \end{array}\right) &=& \left( \begin{array}{r} 0 \\ 2 \\ -6 \end{array}\right) - \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) \\\\ \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) &=& \left( \begin{array}{r} 0 \\ 2 \\ -6 \end{array}\right) - \left( \begin{array}{r} 2 \\ 4 \\ -5 \end{array}\right) \\\\ \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) &=& \left( \begin{array}{r} 0-2 \\ 2-4 \\ -6-(-5) \end{array}\right) \\\\ \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) &=& \left( \begin{array}{r} 0-2 \\ 2-4 \\ -6+5 \end{array}\right) \\\\ \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) &=& \left( \begin{array}{r} -2 \\ -2 \\ -1 \end{array}\right) \\ \hline \end{array}$$

$$\mathbf{\vec{D} =} \left( \begin{array}{r} \mathbf{-2} \\ \mathbf{-2} \\ \mathbf{-1} \end{array}\right)$$

Jan 23, 2020
#4
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Thank you, Omi and heureka!

Jan 23, 2020