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Let P be the intersection point of the line through points D = (1, 1, 2) and E = (2, 3, 4) with the plane through A = (0,1,1), B = (1,1,0) and C = (1,0,3). What is P?

 Aug 5, 2019
 #1
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Let P be the intersection point of the line through points D = (1, 1, 2) and E = (2, 3, 4) 

with the plane through A = (0,1,1), B = (1,1,0) and C = (1,0,3).

What is P?

 

\(\begin{array}{|rcl|rclrcl|} \hline \text{plane} &&& \text{line} \\ \hline && A = \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix} \qquad B = \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} \qquad C = \begin{pmatrix} 1\\ 0\\ 3 \end{pmatrix} \qquad & && D = \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} \qquad E = \begin{pmatrix} 2\\ 3\\ 4 \end{pmatrix} \\ \vec{x} &=& \vec{B}+s(\vec{A}-\vec{B})+t(\vec{C}-\vec{B}) & \vec{x} &=& \vec{D}+r(\vec{E}-\vec{D}) \\\\ \vec{x} &=&\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} +s\left(\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}\right) +t\left(\begin{pmatrix} 1\\ 0\\ 3 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}\right) & \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} +r\left(\begin{pmatrix} 2\\ 3\\ 4 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}\right) \\\\ \vec{x} &=&\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} & \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \vec{x} =\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} &=&\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} -r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} \\\\ \mathbf{ s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} -r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} } &=& \mathbf{ \begin{pmatrix} 0\\ 0\\ 2 \end{pmatrix} } \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & -s-r &=& 0 \\ & \mathbf{r} &=& \mathbf{ -s} \\ \hline (2) & -t-2r &=& 0 \\ & t &=& -2r \\ & t &=& -2(-s) \\ & \mathbf{t} &=& \mathbf{2s} \\ \hline (3) & s+3t-2r &=& 2 \\ & s +3(2s)-2(-s) &=& 2 \\ & s+6s+2s &=& 2 \\ & 9s &=& 2 \\ & \mathbf{s} &=& \mathbf{\dfrac{2}{9}} \\\\ & t &=& 2s \\ & t &=& 2\left(\dfrac{2}{9}\right) \\ &\mathbf{t} &=& \mathbf{\dfrac{4}{9}} \\\\ & r &=& -s \\ & r &=& -\left(\dfrac{2}{9}\right)\\ &\mathbf{r} &=& \mathbf{- \dfrac{2}{9} } \\ \hline \end{array}\)

 

\(\mathbf{\vec{P}=\ ?}\)

\(\begin{array}{|rcll|} \hline \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \vec{P} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}- \dfrac{2}{9}\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \vec{P} &=& \begin{pmatrix} 1-\dfrac{2}{9}\\ 1-\dfrac{4}{9}\\ 2-\dfrac{4}{9} \end{pmatrix} \\\\ \mathbf{\vec{P}} &=& \mathbf{\begin{pmatrix} \dfrac{7}{9}\\ \dfrac{5}{9}\\ \dfrac{14}{9} \end{pmatrix} } \\ \hline \end{array} \)

 

laugh

 Aug 5, 2019

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