# Show that vector area of a parallelogram ABCD is ACBD12(AC¯×BD¯) where AC and BD are its diagonals. - Mathematics and Statistics

Sum

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")

Concept: Vector Product of Vectors (Cross)
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