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

au4gsf · Accepted Answer

We must locate the plane&#039;s normal vector in order to determine its equation.We&#039;ll refer to the indicated positions as P(2,1,2), Q(3,-8,6), and R. (-2,-3,1)Note that&amp;nbsp;PQ→&amp;nbsp;and&amp;nbsp;PR→&amp;nbsp;are vectors in the plane&amp;nbsp;Consequently, their cross product will provide us with the plane&#039;s normal vector.PQ→=(3,−8,6)−(2,1,2)=(3−2,−8−1,6−2)=(1,−9,4)&amp;nbsp;PR→=(−2,−3,1)−(2,1,2)=(−2−2,−3−1,1−2)=(−4,−4,−1)&amp;nbsp;n=PQ→×PR→=|ijk1−94−4−4−1|=(25,−15,−40)&amp;nbsp;Equation of a plane through (a,b,c) with normal vector (l,m,n) isl(x−a)+m(y−b)+n(z−c)=0&amp;nbsp;In the preceding cell, we located the normal vector. We can select any one of the three locations on the plane, so I&#039;ll go with P. (2,1,2)The plane&#039;s equation is25(x−2)−15⋅(y−1)−40⋅(z−2)]=0&amp;nbsp;Divide throughout by 5&amp;nbsp;5(x−2)−3⋅(y−1)−8⋅(z−2)=0&amp;nbsp;5x−3y−8z+9=0&amp;nbsp;Result:&amp;nbsp;5x−3y−8z+9=0

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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