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")]`.

Answer 1

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