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Question
Using properties of scalar triple product, prove that `[(bar"a" + bar"b", bar"b" + bar"c", bar"c" + bar"a")] = 2[(bar"a", bar"b", bar"c")]`.
Solution
L.H.S = `[(bar"a" + bar"b", bar"b" + bar"c", bar"c" + bar"a")]`
`= (bar"a" + bar"b").[(bar"b" + bar"c") xx (bar"c" + bar"a")]`
`= (bar"a" + bar"b").[bar"b" xx bar"c" + bar"b" xx bar"a" + bar"c" xx bar"c" + bar"c" xx bar"a"]`
`= (bar"a" + bar"b").[bar"b" xx bar"c" + bar"b" xx bar"a" + bar"c" xx bar"a"] ....[∵ bar"c" xx bar"c" = bar"0"]`
`= bar"a".[(bar"b" xx bar"c") + (bar"b" xx bar"a") + (bar"c" xx bar"a")] + bar"b".[(bar"b" xx bar"c") + (bar"b" xx bar"a") + (bar"c" xx bar"a")]`
`= bar"a".(bar"b" xx bar"c") + bar"a".(bar"b" xx bar"a") + bar"a".(bar"c" xx bar"a") + bar"b".(bar"b" xx bar"c") + bar"b"(bar"b" xx bar"a") + bar"b"(bar"c" xx bar"a")`
`= [bar"a" bar"b" bar"c"] + [bar"a" bar"b" bar"a"] + [bar"a" bar"c" bar"a"] + [bar"b" bar"b" bar"c"] + [bar"b" bar"b" bar"a"] + [bar"b" bar"c" bar"a"]`
`= [bar"a" bar"b" bar"c"] + 0 + 0 + 0 + 0 + [bar"a" bar"b" bar"c"] `
`= 2[bar"a" bar"b" bar"c"]`
= R.H.S
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