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Calculate a normal vector n to the plane through the points triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2), as in the picture below, such that if

\(\mathbf{n} = \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}, \)

and \(n_1 + n_2 + n_3 = 7. \)

 Aug 17, 2019
 #1
avatar+6046 
+1

\(AB=B-A\\ AC=C-A\\ n = AB \times AC\\ \hat{n} = \dfrac{n}{\|n\|}\\ \text{The scaled version of the normal vector the problem is after is $-7\hat{n}$}\)

 

I leave plugging and chugging to you.

 Aug 17, 2019
 #2
avatar+23884 
+2

Calculate a normal vector n to the plane through the points
triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2), as in the picture below,
such that if
\(\mathbf{n} = \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}\),
and \(n_1 + n_2 + n_3 = 7\).

\(\text{Let $\vec{v} = \vec{B} - \vec{C}$ } \\ \text{Let $\vec{w} = \vec{A} - \vec{C}$ }\)

 

\(\begin{array}{|rcll|} \hline \vec{v} &=& \begin{pmatrix}-3 \\ 1 \\ 4 \end{pmatrix} - \begin{pmatrix}1 \\1\\ 2\end{pmatrix} \\ \mathbf{\vec{v}} &=& \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} \\\\ \vec{w} &=& \begin{pmatrix}1 \\ -2 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\1\\ 2\end{pmatrix} \\ \mathbf{\vec{w}} &=& \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} \\ \hline \end{array}\)

 

1.

Vector cross product

\(\small{ \begin{array}{|lrcll|} \hline & \vec{n} =\begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}&=& \vec{v} \times \vec{w} \\\\ & &=& \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} \times \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} \\\\ & &=& \begin{vmatrix}1&1&1 \\ -4&0&2 \\ 0&-3&-1 \end{vmatrix} \\\\ & &=& \begin{pmatrix} 6 \\ -4 \\ 12 \end{pmatrix} \\ \hline n_1+n_2+n_3: & 6-4+12 &=& 14 \quad | \quad \cdot \dfrac{7}{14} \\\\ & 6\cdot\dfrac{7}{14}-4\cdot\dfrac{7}{14}+12\cdot\dfrac{7}{14} &=& 14\cdot\dfrac{7}{14} \\\\ &\mathbf{ 3-2+6 }&=& \mathbf{7} \\\\ & \vec{n} &=& \begin{pmatrix}\mathbf{3} \\ \mathbf{-2} \\ \mathbf{6} \end{pmatrix} \\ \hline \end{array} }\)

 

2.

Vector dot product

\(\begin{array}{|lrcll|} \hline & \vec{n}\cdot \vec{v} &=& 0 \\ & \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix} \cdot \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} &=& 0 \\ (1)& \mathbf{-4n_1+2n_3} &=& \mathbf{0} \\ \hline & \vec{n}\cdot \vec{w} &=& 0 \\ & \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix} \cdot \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} &=& 0 \\ (2) & \mathbf{-3n_2-n_3} &=& \mathbf{0} \\ \hline (3) & \mathbf{n_1+n_2+n_3} &=& \mathbf{7} \\ \hline \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)& \mathbf{-4n_1+2n_3} &=& \mathbf{0} \\ &2n_3&=&4n_1 \quad | \quad :2 \\ & \mathbf{n_3} &=& \mathbf{2n_1} \\\\ (2) & \mathbf{-3n_2-n_3} &=& \mathbf{0} \\ & \mathbf{n_3} &=& \mathbf{-3n_2} \\\\ & n_3= 2n_1 &=& -3n_2 \\ & 2n_1 &=& -3n_2 \quad | \quad :2 \\ & \mathbf{n_1} &=& \mathbf{-\dfrac{3}{2}n_2} \\\\ (3) & \mathbf{n_1+n_2+n_3} &=& \mathbf{7} \\ & -\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\ & -\dfrac{3}{2}n_2 -2n_2 &=& 7 \quad | \quad \cdot (-1 )\\ & \dfrac{3}{2}n_2 +2n_2 &=& -7 \\ & \dfrac{7}{2}n_2 &=& -7 \quad | \quad \cdot \dfrac{2}{7} \\ & n_2 &=& -7\cdot \dfrac{2}{7} \\ & \mathbf{n_2} &=& \mathbf{-2} \\ \hline & n_3 &=& -3n_2 \\ & n_3 &=& -3\cdot (-2) \\ & \mathbf{n_2} &=& \mathbf{6} \\ \hline & n_1 &=& -\dfrac{3}{2}n_2 \\ & n_1 &=& \left(-\dfrac{3}{2}\right)\cdot(-2) \\ & \mathbf{n_2} &=& \mathbf{3} \\\\ & \mathbf{\vec{n}} &=& \begin{pmatrix}\mathbf{3} \\ \mathbf{-2} \\ \mathbf{6} \end{pmatrix} \\ \hline \end{array}\)

 

laugh

 Aug 17, 2019
edited by heureka  Aug 17, 2019

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