Question

If `vec"a" + vec"b" + vec"c"` = 0, show that `vec"a" xx vec"b" = vec"b" xx vec"c" = vec"c" xx vec"a"`. Interpret the result geometrically?

Answer 1

Given that `vec"a" + vec"b" + vec"c"` = 0

So, `vec"a" xx (vec"a" + vec"b" + vec"c") = vec"a" xx 0`

⇒ `vec"a" xx vec"a" + vec"a" xx vec"b" + vec"a" xx vec"c"` = 0

⇒ `vec"0" + vec"a" xx vec"b" + vec"a" xx vec"c"` = 0  ....`(vec"a" xx vec"a" = 0)`

⇒ `vec"a" xx vec"b" - vec"c" xx vec"a"` = 0  ....`(vec"a" xx vec"c" = -vec"c" xx vec"a")`

⇒ `vec"a" xx vec"b" = vec"c" xx vec"a"`  .....(i)

Now `vec"a" + vec"b" + vec"c"` = 0

⇒ `vec"b" xx (vec"a" + vec"b" + vec"c") = vec"b" xx 0`

⇒ `vec"b" xx vec"a" + vec"b" xx vec"b" xx vec"c"` = 0

⇒ `vec"b" xx vec"a" + vec0 + vec"b" xx vec"c"` = 0  ....`(because vec"b" xx vec"b" = 0)`

⇒ `-(vec"a" xx "b") + vec"b" xx vec"c"` = 0

∴ `vec"b" xx vec"c" = vec"a" xx vec"b"`  ....(ii)

From equation (i) and (ii) we get

`vec"a" xx vec"b" = vec"b" xx vec"c" = vec"c" xx vec"a"`.

Hence proved.

Geometrical Interpretation

According to figure, we have

Area of parallelogram ABCD is 

⇒ `|vec"a" xx vec"b"| = |vec"a"||vec"b"| sin theta`

Since, the parallelograms on the same base and between the same parallel lines are equal in area

∴ `|vec"a" xx vec"b"| = |vec"b" xx vec"c"| = |vec"c" xx vec"a"|`

⇒ `vec"a" xx vec"b" xx vec"c" = vec"c" xx vec"a"`.