Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane
The given plane passes through the point ( 1 , 2 , 3 ) and is perpendicular to the plane r → ⋅ ( i ^ + 2 j ^ - 5 k ^ ) + 9 = 0
The position vector of a point ( x , y , z ) is given as,
r → = x i ^ + y j ^ + z k ^
So, the position vector of the point, ( 1 , 2 , 3 ) is,
r → = i ^ + 2 j ^ + 3 k ^
The direction ratios of the normal to the plane, r → × ( i ^ + 2 j ^ − 5 k ^ ) + 9 = 0 are 1, 2 and − 5 and the normal vector is,
N → = ( i ^ + 2 j ^ − 5 k ^ )
The equation of a line passing through a point and perpendicular to the given plane is given by,
l → = r → + λ N → , λ ∈ R .(1)
Substitute values of N → and r → in equation (1),
l → = i ^ + 2 j ^ + 3 k ^ + λ ( i ^ + 2 j ^ - 5 k ^ )
Thus, the equation of a plane passing through the point ( 1 , 2 , 3 ) and perpendicular to the plane r → × ( i ^ + 2 j ^ − 5 k ^ ) + 9 = 0 is l → = i ^ + 2 j ^ + 3 k ^ + λ ( i ^ + 2 j ^ − 5 k ^ ) .
The given plane passes through the point
The position vector of a point
So, the position vector of the point,
The direction ratios of the normal to the plane,
The equation of a line passing through a point and perpendicular to the given plane is given by,
Substitute values of
Thus, the equation of a plane passing through the point
