Question

Show that vector area of a parallelogram ABCD is `1/2 (bar"AC" xx bar"BD")` where AC and BD are its diagonals.

Answer 1

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")`