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)?

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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.!