Q. 3

# 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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