Q. 45.0( 1 Vote )
Find the vector and Cartesian equations of a plane which is at a distance of 6 units from the origin and which has a normal with direction ratios 2, -1, -2.
d = 6
direction ratios of are (2, -1, -2)
To Find : Equation of plane
1) Unit Vector :
Let be any vector
Then the unit vector of is
2) Dot Product :
If are two vectors
3) Equation of plane :
Equation of plane which is at a distance of 5 units from the origin and having as a unit vector normal to it is
For given normal vector
Unit vector normal to the plane is
Equation of the plane is
This is vector equation of the plane.
= (x × 2) + (y × (-1)) + (z × (-2))
= 2x - y – 2z
Therefore equation of the plane is
This is Cartesian equation of the plane.
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