Question

Find the equation of the plane passing through the three points (2,-1,4),(1,1,1),(3,1,2)

Answer 1

Note that the vectors between these points lie in (i.e. are parallel to) our plane. We can find two of them easily:$$\mathbf{u}=(2,-1,4)-(1,1,1)=(1,-2,3)\\\mathbf{v}=(3,1,2)-(1,1,1)=(2,0,1)$$Now that we have two vectors in the plane, we can find one orthogonal to them using the cross product $$\mathbf{u}\times\mathbf{v}=(-2-0,6-1,0+4)=(-2,5,4)$$Note that this vector is orthogonal to two vectors in the plane and is thus suitable to use as a *normal* to our plane -- every vector lying in said plane will be orthogonal to it.

Now, how do we use all this information? Well recall that we can find a vector in the plane given a point in the plane by finding the vector from a known point (say \((1,1,1)\)) to said point. So if \((x,y,z)\) lies in the plane, we know the vector \((x-1,y-1,z-1)\) lies in it, too. Given the normal, we know every vector will be orthogonal to it (i.e. the two will yield a dot product of \(0\)).$$(x-1,y-1,z-1)\cdot(-2,5,4)=0\\-2(x-1)+5(y-1)+4(z-1)=0\\-2x+2+5y-4+4z-4=0\\-2x+5y+4z-6=0$$