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Question

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

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Solution

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 =( 21 ) i ^ +( 31 ) j ^ +(52) k ^ = i ^ +2 j ^ +3 k ^

BC =( 12 ) i ^ +( 53 ) j ^ +(55) 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 .


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