Parametric form of the equation of the plane is `bar r =(2hati + hatk ) + lambda hati + mu (hati + 2hati-3 hatk) lambda and mu` are parameters. Find normal to the plane and hence equation of the plane in normal form. Write its Cartesion form.

Explanation

The equation `barr = bara + lambdabarb + mubarc` represents a plane passing through a point having position vector a and parallel to vectors `barb` and `barc.`
Here, `bara = 2hati + hatk, barb = hati, barc = hati + 2hatj - 3hatk`
`therefore` The normal vector `barn` to the plane is give by
`barnxxbarbxxbarc=|{:(hati,,hatj,,hatk),(1,,0,,0),(1,,2,,"-3"):}|`
` = (0-0) hati -(-3-0) hatj + (2-0)hatk`
i.e., `barn = 3hatj + 2hatk`
The equation of the plane is scalar product form is
` barr * (barb xx barc) = bara * (barb xx barc)`
i.e., `barr * barn = bara * barn`
where `bara * barn = (2hati + hatk) * (3hati + 2hatk)`
` = 2(0) + 0(3) + 1(2)`
` = 0 + 0 + 2 = 2`
`therefore` the equation of the plane is scalar product form is
`barr*(3hatj + 2hatk) =2`
i.e., `barr * barn = 2`, where `barn = 3hatj + 2hatk`
`therefore |barn| = sqrt(3^(2) + 2^(2)) = sqrt(13)`
The equation `barr * barn = 2` can be written as
`barr*(barn)/(|barn|)=(2)/(|barn|)`
i.e., `barr*((3hatj+2hatk)/(sqrt(13)))=(2)/(sqrt(13))`
i.e., `barr*((3)/(sqrt(13))hatj+(2)/(sqrt(13))hatk)=(2)/(sqrt(13))`
This is the normal form of the equation of plane.
If `barr = xhati + yhatj + zhatk`, then equation (1) becomes
`(xhati+yhatj+zhatk)*(3hatj+2hatk)=2`
`therefore x(0)+y(3)+z(2)=2`

` therefore 3y + 2z = 2`
This is the cartasian equation of the plane.