Question

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

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

2021-06-04
To find the equation of the plane we need to find its normal vector.
Let us call the given points as: P(2,1,2), Q(3,-8,6), R(-2,-3,1)
Note that $$\vec{PQ}$$ and $$\vec{PR}$$ are vectors in the plane
Hence, their cross product will give us normal vector to the plane
$$\vec{PQ}=(3,-8,6)-(2,1,2)=(3-2,-8-1,6-2)=(1,-9,4)$$
$$\vec{PR}=(-2,-3,1)-(2,1,2)=(-2-2,-3-1,1-2)=(-4,-4,-1)$$
$$n=\vec{PQ}\times\vec{PR}=\begin{vmatrix}i&j&k\\1&-9&4\\-4&-4&-1 \end{vmatrix}=(25,-15,-40)$$
Equation of a plane passing through the point (a,b,c) and having normal vector (l,m,n) is
$$l(x-a)+m(y-b)+n(z-c)=0$$
We found the normal vector in the previous cell. For a point on the plane, we can choose any of the three given points, I will choose P(2,1,2)
Equation of the plane is
$$25(x-2)-15\cdot(y-1)-40\cdot(z-2)]=0$$
Divide throughout by 5
$$5(x-2)-3\cdot(y-1)-8\cdot(z-2)=0$$
$$5x-3y-8z+9=0$$
Result:
$$5x-3y-8z+9=0$$