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