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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
avatar+8963 
+1

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
avatar+12269 
+2

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.

laugh

 Jan 23, 2020
 #3
avatar+25275 
+3

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.


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

 

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

 

laugh

 Jan 23, 2020
 #4
avatar+8963 
+2

Thank you, Omi and heureka!

 Jan 23, 2020

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