How do we find out whether four points #A(3,-1,-1),B(-2,1,2)#, #C(8,-3,0)# and #D(0,2,-1)# lie in the same plane or not?

Question

Sophia Alfred · Accepted Answer

#A,B,C# and #D# do not lie in the same plane.

Three non-collinear points are always define a plane. If fourth plane too is on this plane, four plane define this plane. So let us first define a plane using points #A(3,-1,-1),B(-2,1,2)# and #D(0,2,-1)#, using #vec(AB)=(B_x-A_x)hati+(B_y-A_y)hatj+(B_z-A_z)hatk# Therefore #vec(AB)=(-2-3)hati+(1-(-1))hatj+(2-(-1))hatk=# = #-5hati+2hatj+3hatk# and #vec(AD)=(0-3)hati+(2-(-1))hatj+(-1-(-1))hatk=# = #-3hati+3hatj+0hatk# If #vec(AB)# and #vec(AD)# are in the same plane, then we will have #vec(AB)xxvec(AD)=0#, the cross product of the two vector as #0# and hence #|(hati,hatj,hatk),(-5,2,3),(-3,3,0)|=0# or #(0-9)hati-(0-(-9))hatj+(-15-(-6))hatk=0# or #-9hati-9hatj-9hatk=0# or #hati+hatj+hatk=0# Hence equation of plane is #x+y+z=k# and putting values of points #A,B# and #D#, we get #k=1# Hence equation of plane is #x+y+z=1# and as #C(8,-3,0)# does not satisfy it, #A,B,C# and #D# do not lie in the same plane.

Avery Allen · Answer

undefined To determine if the points $ A(3, -1, -1) $, $ B(-2, 1, 2) $, $ C(8, -3, 0) $, and $ D(0, 2, -1) $ lie in the same plane, we can check if the vectors formed by connecting any three of the points are coplanar. If they are coplanar, then all four points are in the same plane. Otherwise, they are not.

We can use the cross product to check coplanarity. If the cross product of two vectors is zero, then they are parallel, and the points lie on the same plane. We'll choose three vectors formed by the points and calculate their cross products:

1. $ \vec{AB} = B - A $
2. $ \vec{AC} = C - A $
3. $ \vec{AD} = D - A $

Then, we'll calculate the cross products of pairs of these vectors:

1. $ \vec{AB} \times \vec{AC} $
2. $ \vec{AB} \times \vec{AD} $

If both cross products are zero, the points are coplanar; if not, they are not coplanar.