Question

Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Answer 1

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 .