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)?

Bruno Schneider

Bruno Schneider

Answered question

2023-03-11

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)?

Answer & Explanation

Jaidyn Velez

Jaidyn Velez

Beginner2023-03-12Added 2 answers

We know that, given two vectors, say x & y , their Vector
or Outer Product , denoted by x × y , is a vector that is
perpendicular to the plane that contains them.
The given pts. A ( 3 , - 1 , 2 ) , B ( 1 , - 1 , - 3 ) and C ( 4 , - 3 , 1 ) lie in the plane A B C
Accordingly, the vectors A B and A C the plane A B C .
Thus, A B × A C plane A B C
Therefore, the reqd. unit vector will be
A B × A C | | A B × A C | |
We have, A B = ( 1 - 3 , - 1 + 1 , - 3 - 2 ) = ( - 2 , 0 , - 5 ) ,
A C = ( 1 , - 2 , - 1 ) , so that,
A B × A C = det | i j k - 2 0 - 5 1 - 2 - 1 |
= - 10 i - 7 j - 4 k = ( - 10 , - 7 , - 4 )
| | A B × A C | | = 100 + 49 + 16 = 165
Finally, the desired unit vector is
( - 10 165 , - 7 165 , - 4 165 ) .

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