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Question
Show that the vector area of a triangle ABC, the position vectors of whose vertices are `bar"a", bar"b" and bar"c"` is `1/2[bar"a" xx bar"b" + bar"b" xx bar"c" + bar"c" xx bar"a"]`.
Solution
Consider the triangle ABC.
Complete the parallelogram ABDC.
Vector area of Δ ABC.
`= 1/2("vector area of parallelogram ABDC")`
`= 1/2(bar"AB" xx bar"AC")`
`= 1/2[(bar"b" - bar"a")xx(bar"c" - bar"a")] ......[∵ bar"AB" = bar"b" - bar"a" and bar"AC" = bar"c" - bar"a"]`
`= 1/2 [bar"b" xx bar"c" - bar"b" xx bar"a" - bar"a" xx bar"c" + bar"a" xx bar"a"]`
`= 1/2 [bar"b"xxbar"c" + bar"a"xxbar"b" + bar"c"xx bar"a" + bar0]`
`= 1/2[bar"a"xxbar"b" + bar"b" xx bar"c" + bar"c" xx bar"a"]`
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