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Question
If `|bar"a"| = |bar"b"| = 1, bar"a".bar"b" = 0, bar"a" + bar"b" + bar"c" = bar"0", "find" |bar"c"|`.
Solution
`bar"a" + bar"b" + bar"c" = bar"0"`
∴ `- bar"c" = bar"a" + bar"b"`
Taking dot product of both sides with itself, we get
`(-bar"c").(-bar"c") = (bar"a" + bar"b").(bar"a" + bar"b")`
∴ `|bar"c"|^2 = bar"a".(bar"a" + bar"b") + bar"b".(bar"a" + bar"b")`
`= bar"a".bar"a" + bar"a".bar"b" + bar"b"bar"a" + bar"b"bar"b"`
`= |bar"a"|^2 + 0 + 0 + |bar"b"|^2 ....[bar"a".bar"b" = bar"b".bar"a" = 0]`
= 1 + 1 = 2 .....`[|bar"a"| = |bar"b"| = 1]`
∴ `|bar"c"| = sqrt2`.
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