Finding Equation of a Plane through the origin and the points$ (1, −2, 5)$ and $(8, 3, 2)$

Question

Find an equation of the plane. The plane through the origin and the points $(1, −2, 5)$ and $(8, 3, 2)$

I know $AB$ is $<7,5,-3>$ but I don't know what to do after that

Answer 1

The vectors $\langle 1,-2,5\rangle$ and $\langle 8,3,2\rangle$ lie in the plane, so the cross product $\vec{n} = \langle 1,-2,5\rangle \times \langle 8,3,2\rangle$ is normal to the plane. Since the plane passes though the origin, the equation of the plane is given by $\vec{n}\cdot \langle x,y,z\rangle = 0$. Simplify $\vec{n}$ so that you write the equation of the plane in the form $ax + by + cz = 0$.

Answer 2

The general equation of a plane passing through origin is $w^T x=0$. $w$ is the direction of the normal to the plane. In your case the normal direction is in the direction of the cross product of the two vectors given, i.e that is the direction of $w$.