Find an equation of the plane. The plane through the points (2, 1, 2), (3, −8, 6), and (−2, −3, 1)

arenceabigns 2021-05-31 Answered
Find an equation of the plane. The plane through the points (2, 1, 2), (3, −8, 6), and (−2, −3, 1)

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Expert Answer

delilnaT
Answered 2021-06-01 Author has 7242 answers

Let the given points be P(2,1,2),Q(3,-8,6), and R(-2,-3,1)
\(\vec{PQ}=\langle3-2,-8-1,6-2\rangle=\langle1,-9,4\rangle\)
\(\vec{PR}=\langle-2-2,-3-1,1-2\rangle=\langle-4,-4,-1\rangle\)
\(\vec{PQ}\times\vec{PR}=\begin{vmatrix}i&j&k\\1&-9&4\\-4&-4&-1\end{vmatrix}=(9+16)i+(-16+1)j+(-4-36)k\)
\(=25i-15j-40k\)
Therefore, the normal vector to the plane is
\(\vec{n}=\langle25,-15,-40\rangle\)
Since, the plane passes through all the three points we can choose any point to find its equation. So, the equation of the plane through the point P(2,1,2) with normal vector \(\vec{n}=\langle25,-15,-40\rangle\) is
\(25(x-2)-15(y-1)-40(z-2)=0\)
\(\Rightarrow 25x-50-15y+15-40z+80=0\)
\(\Rightarrow 25x-15y-40z+45=0\)

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