vectors U, V W are co-planar, determine the Cartesian equation of the Plane II containing the 3 vectors U,V,W?

Question

vectors U, V W are co-planar, determine the Cartesian equation of the Plane II containing the 3 vectors U,V,W passing through point (-2,3,7).
U= 2i-j+k
V= i-j+3k
w= 3i-j-k

Please include full working out/solutions not just answer .
Thank yoo!

Answer 1

normal vector of the plane=
cross product of two vector
cross product of (V,W)=
2*(2i+5j+k)
or
normal vector=
2i+5j+k
if the normal vector =
(ai+bj+ck)
then
plane:
a(x-x0)+b(y-y0)+c(z-z0)=0
thus:
2(x+2)+5(y-3)+(z-7)=0
==>
2x+5y+z=18

there isn't any need of the third vector in the plane
you can find the plane equation with 2 vector in the plane
if you choose another 2 vector the answer would be same

Answer 2

Vectors U, V W are co-planar, determine the Cartesian equation of the Plane II containing the 3 vectors U,V,W passing through point (-2,3,7).
U= 2i-j+k
V= i-j+3k
w= 3i-j-k

This plane is overdescribed. It only takes two non-collinear vectors lieing in a plane and one point to uniquely describe the plane. Let's choose vectors u and v. The normal vector n, to the plane can be determined by taking the cross product of u and v.

n = u X v = <2, -1, 1> X <1, -1, 3> = <-2, -5, -1>

Any non-zero multiple of n is also a normal vector to the plane. Multiply by -1.

n = <2, 5, 1>

Now that we have n and the point P(-2, 3, 7) we can write the equation of the plane.

2(x + 2) + 5(y - 3) + 1(z - 7) = 0
2x + 4 + 5y - 15 + z - 7 = 0
2x + 5y + z - 18 = 0
______

Let's check to make sure the vector w is in the plane. If it is it should be describable as a multiple of u and v. And indeed

2u - v = w

So w is in the plane as expected.