Find the vector, and Cartesian equations of a plane which passes through the point (1, 4, 6) and the normal vector to the plane is ![]()
Given :
A = (1, 4, 6)
![]()
To Find : Equation of plane.
Formulae :
1) Position Vector :
If A is a point having co-ordinates (a1, a2, a3), then its position vector is given by,
![]()
2) Dot Product :
If
are two vectors
![]()
![]()
then,
![]()
3) Equation of plane :
Equation of plane passing through point A and having
as a unit vector normal to it is
![]()
Where, ![]()
Position vector of point A = (1, 4, 6) is
![]()
Now,
![]()
= (1×1) + (4×(-2)) + (6×1)
= 1 – 8 + 6
= - 1
Equation of plane is
![]()
![]()
This is vector equation of the plane.
As ![]()
Therefore
![]()
= (x × 1) + (y × (-2)) + (z × 1)
= x - 2y + z
Therefore equation of the plane is
![]()
This is Cartesian equation of the plane.