Given, vertices of △ABC are A(1,2,3),B(2,3,−1) and C(3,−1,−2) Now, AB=B−A=(2,3,−1)−(1,2,3)=(1,1,−4) Now, CB=(3,−1,−2)−(2,3,−1)=(1,−4,−1) So, |AB|=√12+12+(−4)2 =√1+1+16=√18=3√2 |CB|=√12+(−4)2+(−1)2 =√18=3√2 So, d.c. of AB=
AB
|AB|
=
(1,1,−4)
3√2
=(
1
3√2
,
1
3√2
,
−4
3√2
) And d.c. of CB=
CB
|CB|
=
(1,−4,−1)
3√2
=(
1
3√2
,
−4
3√2
,
−1
3√2
) ∴ d.c.'s of the internal bisector of ∠ABC are (
