Question

Prove by vector method that the parallelograms on the same base and between the same parallels are equal in area

Answer 1

`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.