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)?

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Answer 1 · 2016-08-29T16:06:39

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

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.!