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Question
Prove by vector method that the parallelograms on the same base and between the same parallels are equal in area
Solution
`bar"AB" = bar"a"`
`bar"AD" = bar"b"`
Vector area of the parallelogram is `bar"b" xx bar"a"` ........(1)
Consider the parallelogram ABB’A’
`bar"AB" = bar"a", bar"AB" = "m"bar"a"`
Because `bar"A'B"` is parallel to `bar"AB"`
Consider the triangle ADA’
By law of vectors AA’ = `"m"bar"a" + bar"b"`
Hence the vector area of parallelogram
ABB'A = `bar"a" xx ("m"bar"a" + bar"b")`
= `"m"(bar"a" xx bar"a") + (bar"a" xx bar"b")`
= 0 + `(bar"a" xx bar"b")`
= `bar"a" xx bar"b"` .......(2)
Hence the vector area of parallelogram
By (1) and (2)
Area of ABCD = Area of ABB’A’
Hence proved.
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