Show that the four points A(3, 2, -5),
B(-1, 4, -3), C(-3, 8, -5) and D(-3, 2, 1) are coplanar. Find the equation of the plane containing them.

Given Points :


A = (3, 2, -5)


B = (-1, 4, -3)


C = (-3, 8, -5)


D = (-3, 2, 1)


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 = (3, 2, -5)


B = (-1, 4, -3)


C = (-3, 8, -5)


D = (-3, 2, 1)


Position vectors are given by,






Equation of line passing through points A & B is






Let,


Where,


&


And the equation of the line passing through points C & D is






Let,


Where,


&


Now,





Therefore,



= 72 + 48 – 120


= 0


……… eq(1)


And



= - 72 + 192 – 120


= 0


……… 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,



= 24x + 24y + 24z


From eq(1)



Therefore, equation of required plane is


24x + 24y + 24z = 0


x + y + z = 0


1