Question

Parametric form of the equation of the plane is `bar r=(2hati+hatk)+lambdahati+mu(hat i+2hatj+hatk)` λ and μ are parameters. Find normal to the plane and hence equation of the plane in normal form. Write its Cartesian form.

Answer 1

The vector equation of the plane `barr =bara+lambda bar b+mu barc` in scalar product form is

`barr.(bar b xxbarc)=bara.(barbxxbarc)`

Here , `bar a=2hati+hatk, barb=hati, barc=hati+2hatj+hatk`

`therefore barb xxbarc=|[hati,hatj,hatk],[1,0,0],[1,2,1]|`

`=(0-0)hati-(1-0)hatj+(2-0)hatk`

`=-hatj+2hatk`

 and `bara.(barbxxbarc)=(2hati+hatk).(-hatj+2hatk)`

=2(0)+0(-1)+1(2)=2

the vector equation of the given plane is scalar product form is 

`bar r.(-hatj+2hatk)=2`

If ` barr =xhati+yhatj+zhatk, ` then the above equation becomes,

`(xhati+yhatj+zhatk)(-hatj+2hatk)=2`

`x(0)+y(-1)+z(2)=2`

`∴ -y+2z=2 ` This is the cartesian form of the equation of required plane