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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
Here,
`vec (BC) = (hati + 5hatj + 5hatk) - (2hati + 3hatj + 5hatk)`
`= -hati + 2hatj`
`vec(BA) = (hati + hatj + 2hatk) - (2hati + 3hatj + 5hatk)`
`= -hati - 2hatj - 3hatk`
∴ `vec(BC) xxvec (BA) = abs((hati, hatj, hatk), (-1, 2, 0), (-1, -2, -3))`
`= (-6 + 0)hati - (3 + 0)hatj + (2 + 2)hatk`
`= -6hati - 3hatj + 4hatk`
So, `|vec(BC) xx vec( BA)| = sqrt(36 + 9 + 16)`
`= sqrt61`
∴ Area of `triangle ABC = 1/2 |vec(BC) xx vec(BA)|`
`= 1/2 (sqrt61)` sq. units.
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