How would you find a unit vector perpendicular to plane ABC where points are A(3,-1,2), B(1,-1,-3) and C(4,-3,1)?

Question

54153 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-27T20:47:10

The desired unit vector is

$(-10/sqrt165,-7/sqrt165,-4/sqrt165).$

We know that, given two vectors, say $vecx & vecy$ , their Vector

or Outer Product , denoted by $vecx xx vecy,$ is a vector that is

perpendicular to the plane containing them.

The given pts. $A(3,-1,2), B(1,-1,-3) and C(4,-3,1)$ lie in the

plane $ABC$ .

Accordingly, the vectors $vec(AB) and vec(AC) in" the plane "ABC.$

Hence, $vec(AB) xx vec(AC) bot "plane "ABC$ .

Finally, the reqd. unit vector will be

$(vec(AB) xx vec(AC))/||vec(AB) xx vec(AC)||$

We have, $vec(AB)=(1-3,-1+1,-3-2)=(-2,0,-5)$ ,

$vec(AC)=(1,-2,-1)$ , so that,

$vec(AB) xx vec(AC)=det|(i,j,k),(-2,0,-5),(1,-2,-1)|$

$=-10i-7j-4k=(-10,-7,-4)$

$rArr ||vec(AB) xx vec(AC)|| =sqrt(100+49+16)=sqrt165$ .

Finally, the desired unit vector is

$(-10/sqrt165,-7/sqrt165,-4/sqrt165).$

Enjoy Maths.!