Find an equation of the plane passing through the three points given. P = (2, 0, 0), Q = (0, 4, 0), R = (0, 0, 2)

SchachtN 2021-03-06 Answered

Find an equation of the plane passing through the three points given. \(P = (2, 0, 0),\)

\(Q = (0, 4, 0),\)

\(R = (0, 0, 2)\)

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Expert Answer

broliY
Answered 2021-03-07 Author has 7970 answers

Let \(P=(2,0,0) Q=(0,4,0)\) and \(R=(0,0,2)\)
First, find a normal vector n
\(n=\vec{PQ}\cdot\vec{PR}\)
Therefore,
\(\vec{PQ}=(0,4,0)-(2,0,0)=(-2,4,0)\)
\(\vec{PR}=(0,0,2)-(2,0,0)=(-2,0,2)\)
\(n=\vec{PQ}\cdot\vec{PR}=\begin{bmatrix}i&j&k\\-2&4&0\\-2&0&2\end{bmatrix}\)
\(n=\begin{bmatrix}4&0\\0&2\end{bmatrix}i-\begin{bmatrix}-2&0\\-2&2\end{bmatrix}j+\begin{bmatrix}-2&4\\-2&0\end{bmatrix}k\)
\(n=(8-0)i-(-4+0)j+(0+8)k\)
\(n=8i+4j+8k\)
\(n=(8,4,8)\)
We have the equation \(8x+4y+8z=d\)
Now choose any one of the three points, using \(P=(2,0,0)\)
\(d=n\cdot\vec{OP}\)
\(d=(8,4,8)\cdot(2,0,0)\)
\(d=16\)
\(d=1\)
Therefore
\(8x+4y+8z=16\)
\(2x+y+2z=4\)
Results
\(2x+y+2z=4\)

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