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Question
If `bar"OA" = bar"a" and bar"OB" = bar"b",` then show that the vector along the angle bisector of ∠AOB is given by `bar"d" = lambda(bar"a"/|bar"a"| + bar"b"/|bar"b"|).`
Solution
Choose any point P on the angle bisector of ∠AOB. Draw PM parallel to OB.
∴ ∠OPM = ∠POM = ∠POB
Hence, OM = MP
∴ OM and MP is the same scalar multiple of unit vectors `hat"a" and hat"b"` along these directions,
where `hat"a" = bar"a"/|bar"a"| and hat"b" = bar"b"/|bar"b"|`
∴ `bar"OM" = lambdahat"a" and bar"MP" = lambdahat"b"`
∴ `bar"OP" = bar"OM" + bar"MP"`
`= lambdahat"a" + lambdahat"b"`
`= lambda(hat"a" + hat"b")`
Hence, the vector along angle bisector of ∠AOB is given by
`bar"d" = bar"OP" = lambda(bar"a"/|bar"a"| + bar"b"/|bar"b"|)`.
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