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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