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# math problem

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The equation of the plane through the points triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2) can be written as ax + by + cz = 13. What is the ordered triple      (a, b, c)?

Aug 15, 2019

#1
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The equation of the plane through the points triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2) can be written as ax + by + cz = 13.

What is the ordered triple   (a, b, c)?

Formula:  $$\vec{x} = \vec{C} + r(\vec{A}-\vec{C}) +s(\vec{B}-\vec{C})$$

$$\begin{array}{|lrcll|} \hline & \mathbf{\vec{x}} &=& \mathbf{ \vec{C} + r(\vec{A}-\vec{C}) +s(\vec{B}-\vec{C}) } \\ & \begin{pmatrix} x\\ y\\ z \end{pmatrix} &=&\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} +r\left(\begin{pmatrix} 1\\ -2\\ 1 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}\right) +s\left(\begin{pmatrix} -2\\ 1\\ 4 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}\right) \\ & \begin{pmatrix} x\\ y\\ z \end{pmatrix} &=&\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} +r \begin{pmatrix} 0\\ -3\\ -1 \end{pmatrix} +s \begin{pmatrix} -4\\ 0\\ 2 \end{pmatrix} \\ \hline (1) & x &=& 1-4s \\ & 4s &=& 1-x \\ & \mathbf{s} &=& \mathbf{\dfrac{1}{4}(1-x)} \\\\ (2) & y &=& 1-3r \\ & 3r &=& 1-y \\ & \mathbf{r} &=& \mathbf{\dfrac{1}{3}(1-y)} \\\\ (3) & z &=& 2-r+2s \\ & z &=& 2-\dfrac{1}{3}(1-y)+\dfrac{2}{4}(1-x) \\ & z &=& 2-\dfrac{1}{3}(1-y)+\dfrac{1}{2}(1-x) \quad | \quad \cdot 6 \\ & 6z &=& 12-\dfrac{6}{3}(1-y)+\dfrac{6}{2}(1-x) \\ & 6z &=& 12-2(1-y)+3(1-x) \\ & 6z &=& 12-2+2y+3-3x \\ & \mathbf{3x-2y+6z } &=& \mathbf{13} \\ \hline \end{array}$$

$$(a,\ b,\ c) = (3,\ -2,\ 6)$$ Aug 15, 2019
#2
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If you are familiar with matrix solutions of simultaneous linear equations, the following matrix solution is an alternative way of tackling this problem: (NB  I've deliberately left out the detail and just sketched the approach for those who are familiar with matrix manipulations.)

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Aug 15, 2019
edited by Alan  Aug 15, 2019