For each of the following planes, find the direction cosines of the normal to the plane and the distance of the plane from the origin:
(i) 2x + 3y - z = 5
(ii) z = 3
(iii) 3y + 5 = 0
(i) 2x + 3y – z = 5
Given :
Equation of plane : 2x + 3y – z = 5
To Find :
Direction cosines of the normal i.e.
Distance of the plane from the origin = d
Formulae :
1) Direction cosines :
If a, b & c are direction ratios of the vector then its direction cosines are given by
2) The distance of the plane from the origin :
Distance of the plane from the origin is given by,
For the given equation of plane
2x + 3y – z = 5
Direction ratios of normal vector are (2, 3, -1)
Therefore, equation of normal vector is
Therefore, direction cosines are
Now, the distance of the plane from the origin is
(ii) Given :
Equation of plane : z = 3
To Find :
Direction cosines of the normal, i.e.
The distance of the plane from the origin = d
Formulae :
3) Direction cosines :
If a, b & c are direction ratios of the vector, then its direction cosines are given by
4) The distance of the plane from the origin :
Distance of the plane from the origin is given by,
For the given equation of a plane
z = 3
Direction ratios of normal vector are (0, 0, 1)
Therefore, equation of normal vector is
Therefore, direction cosines are
Now, the distance of the plane from the origin is
(iii) Given :
Equation of plane : 3y + 5 = 0
To Find :
Direction cosines of the normal, i.e.
The distance of the plane from the origin = d
Formulae :
1) Direction cosines :
If a, b & c are direction ratios of the vector, then its direction cosines are given by
2) Distance of the plane from the origin :
Distance of the plane from the origin is given by,
For the given equation of a plane
3y + 5 = 0
⇒-3y = 5
Direction ratios of normal vector are (0, -3, 0)
Therefore, equation of normal vector is
Therefore, direction cosines are
Now, distance of the plane from the origin is