Using vector method, find the incenter of the triangle whose vertices are A(0, 3, 0), B(0, 0, 4) and C(0, 3, 4)

#### Solution

Let `bar"a", bar"b", bar"c"` be the position vectors of points A, B, C respectively of ∆ABC and `bar"h"` be the position vector of its incentre H.

∴ `bar"h" = (|bar"BC"|bar"a" + |bar"AC"|bar"b" + |bar"AB"|bar"c")/(|bar"BC"| + |bar"AC"| + |bar"AB"|)` ...(i)

∴ `bar"a" = 3hat"j", bar"b" = 4hat"k", bar"c" = 3hat"j" + 4hat"k"`

∴ `bar"BC" = bar"c" - bar"b" = (3hat"j" + 4hat"k") - 4hat"k" = 3hat"j"`

`bar"AC" = bar"c" - bar"a" = (3hat"j" + 4hat"k") - 3hat"j" = 4hat"k"`

`bar"AB" = bar"b" - bar"a" = 4hat"k" - 3hat"j"`

∴ `|bar"BC"| = sqrt(9)` = 3

`|bar"AC"| = sqrt(16)` = 4

and

`|bar"AB"| = sqrt(16 + 9) = sqrt(25)` = 5

∴ `bar"h" = (3(3hat"j") + 4(4hat"k") + 5(3hat"j" + 4hat"k"))/(3 + 4 + 5)` .......[From (i)]

∴ `bar"h" = (9hat"j" + 16hat"k" + 15hat"j" + 20hat"k")/12`

= `(24hat"j" + 36hat"k")/12`

= `2hat"j" + 3hat"k"`

∴ Incentre of the triangle is H (0, 2, 3).