Question

If `bar("c") = 3bar("a") - 2bar("b")` then prove that `[(bar("a"), bar("b"), bar("c"))]` = 0

Answer 1

`bar("c") = 3bar("a") - 2bar("b")`    .......[Given]

`[(bar("a"), bar("b"), bar("c"))] = bar("a")*(bar("b") xx bar("c"))`

= `bar("a")*[bar("b") xx (3bar("a") - 2bar("b"))]`

= `bar("a")*[bar("b") xx 3bar("a") - bar("b") xx 2bar("b")]`

= `bar("a")*[bar("b") xx 3bar("a") - bar(0)]`    .......`[∵  bar("b") xx bar("b") = bar(0)]`

= `3bar("a")*[bar("b") xx bar("a")]`

= `3[(bar("a"), bar("b"), bar("a"))]`

The scalar triple product `bara * (barb xx bara)` is always zero because `(barb xx bara)` is perpendicular to `bara`, nd the dot product of a vector with a perpendicular vector is zero.

`therefore [bara, barb, barc]`= 3(0)

∴ `[(bar("a"), bar("b"), bar("c"))]` = 0