Find the direction cosines of the unit vector perpendicular to the plane ` -> vecr*(6 hat i-3 hat j-2 hat k)+1=0`passing through the origin.

Explanation

It is given that the equation of the plane is,

`vecr*(6 hat i-3 hat j-2 hat k)+1=0`

Here, `vecN=(6 hat i-3 hat j-2 hat k)`

So, `abs(vecN)=sqrt(6^2+(-3)^2+(-2)^2)`

`=>abs(vecN)=sqrt(36+9+4)`

`=>abs(vecN)=sqrt(49)`

`=>abs(vecN)=7`

In order to find the direction cosines of the unit vector perpendicular to the given plane passing through the origin, the given equation must be converted into its normal form as follows,

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