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Find the plane that goes through the points $\left(3, 0, -4\right)$, $\left(-2, -2, 5\right)$, and $\left(-1, 8, -3\right)$.
in Mathematics by Diamond (75,918 points) | 34 views

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Find the plane that goes through the points $\left(3, 0, -4\right)$, $\left(-2, -2, 5\right)$, and $\left(-1, 8, -3\right)$.
\begin{align*}
P &= \left(3, 0, -4\right) && \text{Define points}\\
Q &= \left(-2, -2, 5\right)\\
R &= \left(-1, 8, -3\right)\\\\
\overrightarrow{PQ} &= \left\langle -5, -2, 9 \right\rangle && \text{Vector through points}\\
\overrightarrow{PR} &= \left\langle -4, 8, 1 \right\rangle\\\\
\overrightarrow{PQ} \times \overrightarrow{PR} &= \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{-5}&{-2}&{9}\\{-4}&{8}&{1}\end{array}} \right| && \text{Take the cross product}\\
\overrightarrow{PQ} \times \overrightarrow{PR} &= \left| {\begin{array}{*{20}{c}}{-2}&{9}\\{8}&{1}\end{array}} \right|\vec i - \left| {\begin{array}{*{20}{c}}{-5}&{9}\\{-4}&{1}\end{array}} \right|\vec j + \left| {\begin{array}{*{20}{c}}{-5}&{-2}\\{-4}&{8}\end{array}} \right|\vec k\\\\
\overrightarrow{PQ} \times \overrightarrow{PR} &= \langle -74, -31, -48 \rangle && \text{Normal vector}\\
P &= \left(3, 0, -4\right) && \text{Point on plane}\\\\
0 &= \vec{n}\cdot (r - r_0) && \text{Equation of a plane}\\
&&&\text{$\vec n$: the normal vector}\\
&&&\text{$r_0$: the point on the plane} \\\\
0 &= \langle -74, -31, -48 \rangle \cdot \left(\langle x, y, z\rangle - \langle 3,0,5 \right\rangle) && \text{Equation of the plane}\\
0 &= -74(x-3) - 31y - 48(z-4) && \text{Simplified to scalar notation}
\end{align*}
by Diamond (75,918 points)

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