Question

How do you find a unit vector normal to the plane which passes through the following points (1,1,1),(2,0,2),(1,1,2)?

Answer 1

Reqd. unit normal is `(-1/sqrt2,-1/sqrt2,0), or, (1/sqrt2,1/sqrt2,0)`.

Explanation:

Let `A(1,1,1), B(2,0,2), and, C(1,1,2)` be the pts. on the plane, say `pi`.

We need a unit normal, call it `hat n`, to `pi`.

`vec(AB)=(2-1,0-1,2-1)=(1,-1,1) in pi`, and,

`vec(AC)=(1-1,1-1,2-1)=(0,0,1) in pi`.

`:.` Fom Vector Geometry, `vec(AB) xx vec(AC)` is a vector `bot`

to `vec(AB) and vec(AC)`, hence, is the normal to the plane `pi`.

Now, `vec(AB)xxvec(AC)=det|(hati,hatj,hatk),(1,-1,1),(0,0,1)|`

`=-1hati-1hatj+0hatk=(-1,-1,0)`, so that,

`||vec(AB)xxvec(AC)||=sqrt2`.

Therefore,

`hatn=[vec(AB)xxvec(AC)]/||vec(AB)xxvec(AC)||`, i.e.,

`hat n=(-1/sqrt2,-1/sqrt2,0), or, (1/sqrt2,1/sqrt2,0)`

Enjoy maths.!