Q. 114.5( 2 Votes )

For each of the f

Answer :

(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




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