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Question
If `bar"c" = 3bar"a" - 2bar"b"`, then prove that `[bar"a" bar"b" bar"c"] = 0`.
Solution
We use the results: `bar"b" xx bar"b" = bar"0"` and if in a scalar triple product, two vectors are equal, then the scalar triple product is zero.
`[bar"a" bar"b" bar"c"] = bar"a".(bar"b" xx bar"c")`
`= bar"a".[bar"b" xx (3bar"a" - 2bar"b")]`
`= bar"a".(3bar"b" xx bar"a" - 2bar"b" xx bar"b")`
`= bar"a". (3bar"b" xx bar"a" - bar"0")`
`= 3bar"a".(bar"b" xx bar"a")`
= 3 × 0
= 0
Alternative Method:
`bar"c" = 3bar"a" - 2bar"b"`
∴ `bar"c"` is a linear combination of `bar"a" "and" bar"b"`.
∴ `bar"a" , bar"b" , bar"c"` are coplanar.
∴ `[bar"a" bar"b" bar"c"] = 0`
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