Question

find the d.cs of the normal to the plane x+2y+2z+2=0​

Answer 1

Answer :

⅓ , ⅔ , ⅔

Note :

→ The general equation of a plane is given as ; ax + by + cz + d = 0 .

→ a , b , c are the direction ratios of the normal to the plane .

→ a/√(a² + b² + c²) , b/√(a² + b² + c²) , c/√(a² + b² + c²) are the direction cosines of the normal to the plane .

Solution :

Here ,

The given equation of the plane is ;

x + 2y + 2z + 2 = 0 .

Now ,

Comparing the given equation of the plane with the general equation of the plane ax + bx + cx + d = 0 , we have ;

a = 1

b = 2

c = 2

d = 2

Now ,

=> a² + b² + c² = 1² + 2² + 2²

=> a² + b² + c² = 1 + 4 + 4

=> a² + b² + c² = 9

=> √(a² + b² + c²) = √9

=> √(a² + b² + c²) = 3

Thus ,

• a/√(a² + b² + c²) = 1/3

• b/√(a² + b² + c²) = 2/3

• c/√(a² + b² + c²) = 2/3

Hence ,

⅓ , ⅔ , ⅔ are the direction cosines of the given plane .