Write the equation of a plane passing through the point (2, -1, 1) and parallel to the plane 3x + 2y - z = 7.
Given :
A ≡ (2, -1, 1)
Plane parallel to the required plane : 3x + 2y – z = 7
To Find : Equation of plane
Formulae :
1) Position vectors :
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 :
If a plane is passing through point A, then equation of plane is
Where,
Answer :
For point A ≡ (2, -1, 1), position vector is
As required plane is parallel to 3x + 2y – z = 7.
Therefore, normal vector of given plane is also perpendicular to required plane
Now,
= 6 – 2 – 1
= 3
Equation of the plane passing through point A and perpendicular to vector is
As
= 3x + 2y – z
Therefore, equation of the plane is
3x + 2y – z = 3
3x + 2y – z - 3 = 0