Show that the four points A(0, -1, 0),
B(2, 1, -1), C(1, 1, 1) and D(3, 3, 0) are coplanar. Find the equation of the plane containing them.

Given Points :


A = (0, -1, 0)


B = (2, 1, -1)


C = (1, 1, 1)


D = (3, 3, 0)


To Prove : Points A, B, C & D are coplanar.


To Find : Equation of plane passing through points A, B, C & D.


Formulae :


1) Position vectors :


If A is a point having co-ordinates (a1, a2, a3), then its position vector is given by,



2) Equation of line


If A and B are two points having position vectors then equation of line passing through two points is given by,



3) Cross Product :


If are two vectors




then,



4) Dot Product :


If are two vectors




then,



5) Coplanarity of two lines :


If two lines are coplanar then



6) Equation of plane :


If two lines are coplanar then equation of the plane containing them is



Where,



For given points,


A = (0, -1, 0)


B = (2, 1, -1)


C = (1, 1, 1)


D = (3, 3, 0)


Position vectors are given by,






Equation of line passing through points A & D is






Let,


Where,


&


And equation of line passing through points B & C is






Let,


Where,


&


Now,





Therefore,



= 0 + 6 + 0


= 6


……… eq(1)


And



= 16 – 6 – 4


= 6


……… eq(2)


From eq(1) and eq(2)



Hence lines are coplanar


Therefore, points A, B, C & D are also coplanar.


As lines are coplanar therefore equation of the plane passing through two lines containing four given points is



Now,



= 8x - 6y + 4z


From eq(1)



Therefore, equation of required plane is


8x - 6y + 4z = 6


4x - 3y + 2z = 3


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