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