The given vertices of triangle are A( 1,1,2 ),B( 2,3,5 )and C( 1,5,5 ).
Area of ABC is given by,
1 2 | AB → × B → C |(1)
The position vectors can be formulated as,
AB → =( 2−1 ) i ^ +( 3−1 ) j ^ +(5−2) k ^ = i ^ +2 j ^ +3 k ^
BC → =( 1−2 ) i ^ +( 5−3 ) j ^ +(5−5) k ^ =− i ^ +2 j ^
The cross product of two vectors ( a 1 i ^ + a 2 j ^ + a 3 k ^ ) and ( b 1 i ^ + b 2 j ^ + b 3 k ^ ) is given by,
a → × b= → | i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 |
AB → × BC → =| i ^ j ^ k ^ 1 2 3 −1 2 0 | = i ^ ( −6 )− j ^ ( 3 )+ k ^ ( 2+2 ) =−6 i ^ − j ^ ( 3 )+ k ^ ( 4 )
| AB → × B → C |= ( −6 ) 2 + ( −3 ) 2 + 4 2 = 36+9+16 = 61 (2)
Substitute value of (2) in (1),
1 2 | AB → × B → C |= 61 2
Thus, area of triangle with vertices A( 1,1,2 ),B( 2,3,5 )and C( 1,5,5 )is 61 2 .