Question

Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector  `2hati + hatj + 2hatk.`

Answer 1

The plane is perpendicular to ` barr=2hati + hatj + 2hatk`

the normal vector `barn` to the plane is

`barn=2hati + hatj + 2hatk`

∴ unit vector along this normal is 

`bar n=barn/(|barn|)=(2hati + hatj + 2hatk)/sqrt(2^2+1^2+2^2)`

`=(2hati + hatj + 2hatk)/3`

 The vector equation of the plane in normal form is ` bar r=barn = p` where p is the distance of the plane from the origin. Here p = 5.

`barr=(2hati + hatj + 2hatk)/3=5`

`therefore barr=(2hati + hatj + 2hatk)=15`