Question

$A(5,4,6),B(1,-1,3),C(4,3,2)$ are three points. Find the coordinates of the point in which the bisector of $\angle BAC$ meets the side $\overline{BC}$

Answer 1

A(5,4,6), B(1.-1,3) and C(4,3,2) are the vertices of triangle ABC

Using distance formula

$AB=\sqrt{(1-5{)}^2+(-1-4{)}^2+(3-6{)}^2}$

$=\sqrt{16+25+9}$

$=\sqrt{50}$

$=5\sqrt{2}$

$AC=\sqrt{(4-5{)}^2+(3-4{)}^2+(2-6{)}^2}$

$=\sqrt{1+1+16}$

$=\sqrt{18}$

$=3\sqrt{2}$

Since bisector of $\angle BAC$ meets BC in D

AD is the bisector of $\angle BAC$ we have

$\frac{BD}{DC}=\frac{AB}{AC}=\frac{5\sqrt{2}}{3\sqrt{2}}=\frac{5}{3}$

D divides BC in the ratio 5:3

Hence the coordinates of D $=(\frac{20+3}{8},\frac{15-3}{8},\frac{10+9}{8})$

$=(\frac{23}{8},\frac{12}{8},\frac{19}{8})$