Question

$A(3,2,0),B(5,3,2),C(-9,6,-3)$ are three points forming a triangle and $AD$ is the bisectors of the $\angle BAC$ meet BC at D, then co-ordinates of $D$ are:

• A. $(\frac{17}{16},\frac{57}{16},\frac{28}{16})$
• B. $(\frac{38}{16},\frac{57}{16},\frac{17}{16})$
• C. $(\frac{38}{16},\frac{17}{16},\frac{57}{166})$
• D. $(\frac{57}{16},\frac{38}{16},\frac{17}{16})$

Answer 1

The correct option is B $(\frac{38}{16},\frac{57}{16},\frac{17}{16})$


Note that

$AB=\sqrt{(5-3{)}^2+(3-2{)}^2+(2-0{)}^2}=\sqrt{4+1+4}=3$

$AC=\sqrt{(-9-3{)}^2+(6-2{)}^2+(-3-0{)}^2}=\sqrt{144+16+9}=\sqrt{169}=13$

The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides.

Since, AD is bisector of $\angle BAC$
We have, $\frac{AB}{AC}=\frac{BD}{DC}=\frac{3}{13}$

Now, Point D divides BC in the ratio of 3:13

Hence, Coordinated of D are: $(\frac{-9\left(3\right)+(13\times 5)}{3+13},\frac{(6\times 3)+(13\times 3)}{3+13},\frac{(-3\times 3)+(13\times 2)}{3+13})$

$\Rightarrow (\frac{-27+65}{16},\frac{18+39}{16},\frac{-9+26}{16})$

$\Rightarrow (\frac{38}{16},\frac{57}{16},\frac{17}{16})$
Thus, Option (B) is correct answer.