# 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  2i + j + 2k. - Mathematics and Statistics

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.

#### Solution

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`

Concept: Vector and Cartesian Equation of a Plane
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