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

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

2021-06-01

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
$$25(x-2)-15(y-1)-40(z-2)=0$$
$$\Rightarrow 25x-50-15y+15-40z+80=0$$
$$\Rightarrow 25x-15y-40z+45=0$$