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"|).`

Answer 1

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"|)`.