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

Equation of the plane is
`bar r = ( 2hati + hatk)+ lambda hati + mu ( hati + 2hatj-3hatk)" …(i)"`
Comparing equation (i) With `bar r = bara + lambda barb+ mu barc`
`veca=2(hati +hatk), vecb=hati, vec c= hati + 2hatj - 3hatk`
`therefore ` Normal to the plane is
`vecb xx vecc = |(hati, hatj, hatk),(1,0,0),(1,2,-3)|`
`=hati (0-0)- hatj(-3-0)+hatk(2-0)`
`3hatj+ 2hatk`
Now, `bara. (barb xx barc) = (2hati + hatk). (3 hatj+ 2hatk)`
`=0+0+2=2`
`therefore` Equation of the plane in normal form is
`vec r . (vecb xx vecc)= veca .(vecb xx vecc)`
`vecr. (3hatj+2hatk)=2" ...(ii)"`
If `vec r = xhati + yhatj+ zhatk`, then the equation becomes,
`(x hati+y hatj+ zhatk) . (3hatj+2hatk)=2`
`rArr" "x(0)+y(3)+z(2)=2`
`rArr" "3y+2x=2`
This is the Cartesian form of the equation of given plane.