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Question
Show that vector area of a parallelogram ABCD is `1/2 (bar"AC" xx bar"BD")` where AC and BD are its diagonals.
Solution
Let ABCD be a parallelogram.
Then `bar"AC" = bar"AB" + bar"BC"` and
`bar"BD" = bar"BA" + bar"AD" = - bar"AB" + bar"BC" ....[because bar"BC" = bar"AD"]`
`= bar"BC" - bar"AB"`
∴ `bar"AC" xx bar"BD" = (bar"AB" + bar"BC") xx (bar"BC" - bar"AB")`
`= bar"AB" xx (bar"BC" - bar"AB") + bar"BC" xx (bar"BC" - bar"AB")`
`= bar"AB" xx bar"BC" - bar"AB" xx bar"AB" + bar"BC" xx bar"BC" - bar"BC" xx bar"AB"`
`= bar"AB" xx bar"BC" + bar"AB" xx bar"BC"`
....`[bar"AB" xx bar"AB" = bar"BC" xx bar"BC" = bar"0" "and" - bar"BC" xx bar"AB" = bar"AB" xx bar"BC"]`
∴ `bar"AC" xx bar"BD" = 2(bar"AB" xx bar"BC")`
= 2 (vector area of parallelogram ABCD)
∴ vector area of parallelogram ABCD `= 1/2(bar"AC" xx bar"BD")`
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